[php] INSERT and immediately UPDATE

[Allen McCabe]

Hey everyone, I have an issue.

I need my (employee) users to be able to insert shows into the our MySQL
database and simultaneously upload an image file (and store the path in the
table).

I have accomplished this with a product-based system (adding products and
uploading images of the product), and accomplished what I needed because the
product name was unique; I used the following statements:

$prodName = $_POST['prodName'];
$prodDesc = $_POST['prodDesc'];
$prodPrice = $_POST['prodPrice'];

$query2  = "INSERT INTO product (pID, pName) VALUES (NULL, '$prodName');";
$result2 = mysql_query($query2) or die(mysql_error());

$query  = "SELECT pID FROM product WHERE pName = '$prodName';";
$result = mysql_query($query) or die(mysql_error());
$row = mysql_fetch_array($result) or die (mysql_error());

$prodID = $row['pID'];


I had to select the new product to get the product id to use in the new
unique image name.

The problem I am facing now, is that with the shows that my users add will
have multitple show times; this means non-unique titles. In fact, the only
unique identifier is the show id. How can I insert something (leaving the
show_id field NULL so that it is auto-assigned the next ID number), and then
immediately select it?

PHP doesn't seem to be able to immediately select something it has just
inserted, perhaps it needs time to process the database update.

Here is the code I have now (which does not work):

$query2  = "INSERT INTO afy_show (show_id, show_title, show_day_w,
show_month, show_day_m, show_year, show_time, show_price, show_description,
show_comments_1, show_seats_reqd) VALUES (NULL, '{$show_title}',
'{$show_day_w}', '{$show_month}', '{$show_day_m}', '{$show_year}',
'{$show_time}', '{$show_price}', '{$show_description}',
'{$show_comments_1}', '{$show_seats_reqd}');";
 $result2 = mysql_query($query2) or die(mysql_error());

 $query3 = "SELECT * FROM afy_show WHERE *show_id = '$id'*;";
 $result3 = mysql_query($query3) or die('Record cannot be located!' .
mysql_error());
 $row3 = mysql_fetch_array($result3);
 $show_id = $row3['show_id'];

How do I select the item I just inserted to obtain the ID number??

[Ashley Sheridan-3]

On Wed, 2009-10-28 at 12:21 -0700, Allen McCabe wrote:

> Hey everyone, I have an issue.
>
> I need my (employee) users to be able to insert shows into the our MySQL
> database and simultaneously upload an image file (and store the path in the
> table).
>
> I have accomplished this with a product-based system (adding products and
> uploading images of the product), and accomplished what I needed because the
> product name was unique; I used the following statements:
>
> $prodName = $_POST['prodName'];
> $prodDesc = $_POST['prodDesc'];
> $prodPrice = $_POST['prodPrice'];
>
> $query2  = "INSERT INTO product (pID, pName) VALUES (NULL, '$prodName');";
> $result2 = mysql_query($query2) or die(mysql_error());
>
> $query  = "SELECT pID FROM product WHERE pName = '$prodName';";
> $result = mysql_query($query) or die(mysql_error());
> $row = mysql_fetch_array($result) or die (mysql_error());
>
> $prodID = $row['pID'];
>
>
> I had to select the new product to get the product id to use in the new
> unique image name.
>
> The problem I am facing now, is that with the shows that my users add will
> have multitple show times; this means non-unique titles. In fact, the only
> unique identifier is the show id. How can I insert something (leaving the
> show_id field NULL so that it is auto-assigned the next ID number), and then
> immediately select it?
>
> PHP doesn't seem to be able to immediately select something it has just
> inserted, perhaps it needs time to process the database update.
>
> Here is the code I have now (which does not work):
>
> $query2  = "INSERT INTO afy_show (show_id, show_title, show_day_w,
> show_month, show_day_m, show_year, show_time, show_price, show_description,
> show_comments_1, show_seats_reqd) VALUES (NULL, '{$show_title}',
> '{$show_day_w}', '{$show_month}', '{$show_day_m}', '{$show_year}',
> '{$show_time}', '{$show_price}', '{$show_description}',
> '{$show_comments_1}', '{$show_seats_reqd}');";
>  $result2 = mysql_query($query2) or die(mysql_error());
>
>  $query3 = "SELECT * FROM afy_show WHERE *show_id = '$id'*;";
>  $result3 = mysql_query($query3) or die('Record cannot be located!' .
> mysql_error());
>  $row3 = mysql_fetch_array($result3);
>  $show_id = $row3['show_id'];
>
> How do I select the item I just inserted to obtain the ID number??

Have a look at mysql_insert_id() which returns the auto insert id value
of the last row to be inserted by the current script. If you use if
right after the insert query (without any other queries in between) then
you have the id for that row. Whatever you do, don't look for the
MAX(id) value! I've seen people do this before, and then wonder why the
database gets all corrupted when more than one person uses the system at
the same time!

Thanks,
Ash
http://www.ashleysheridan.co.uk

[João Cândido de Souza Neto]

Use mysql_insert_id() instead of the select you´re using now.

"Allen McCabe" <allenmccabe@...> escreveu na mensagem
news:657acef20910281221y5ab6ab7t4882f4f00da2cd19@......

> Hey everyone, I have an issue.
>
> I need my (employee) users to be able to insert shows into the our MySQL
> database and simultaneously upload an image file (and store the path in
> the
> table).
>
> I have accomplished this with a product-based system (adding products and
> uploading images of the product), and accomplished what I needed because
> the
> product name was unique; I used the following statements:
>
> $prodName = $_POST['prodName'];
> $prodDesc = $_POST['prodDesc'];
> $prodPrice = $_POST['prodPrice'];
>
> $query2  = "INSERT INTO product (pID, pName) VALUES (NULL, '$prodName');";
> $result2 = mysql_query($query2) or die(mysql_error());
>
> $query  = "SELECT pID FROM product WHERE pName = '$prodName';";
> $result = mysql_query($query) or die(mysql_error());
> $row = mysql_fetch_array($result) or die (mysql_error());
>
> $prodID = $row['pID'];
>
>
> I had to select the new product to get the product id to use in the new
> unique image name.
>
> The problem I am facing now, is that with the shows that my users add will
> have multitple show times; this means non-unique titles. In fact, the only
> unique identifier is the show id. How can I insert something (leaving the
> show_id field NULL so that it is auto-assigned the next ID number), and
> then
> immediately select it?
>
> PHP doesn't seem to be able to immediately select something it has just
> inserted, perhaps it needs time to process the database update.
>
> Here is the code I have now (which does not work):
>
> $query2  = "INSERT INTO afy_show (show_id, show_title, show_day_w,
> show_month, show_day_m, show_year, show_time, show_price,
> show_description,
> show_comments_1, show_seats_reqd) VALUES (NULL, '{$show_title}',
> '{$show_day_w}', '{$show_month}', '{$show_day_m}', '{$show_year}',
> '{$show_time}', '{$show_price}', '{$show_description}',
> '{$show_comments_1}', '{$show_seats_reqd}');";
> $result2 = mysql_query($query2) or die(mysql_error());
>
> $query3 = "SELECT * FROM afy_show WHERE *show_id = '$id'*;";
> $result3 = mysql_query($query3) or die('Record cannot be located!' .
> mysql_error());
> $row3 = mysql_fetch_array($result3);
> $show_id = $row3['show_id'];
>
> How do I select the item I just inserted to obtain the ID number??
>

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[Mari Masuda-2]

Maybe you could use http://us.php.net/manual/en/function.mysql-insert-id.php 
  to get the inserted id.

On Oct 28, 2009, at 12:21 PM, Allen McCabe wrote:

> Hey everyone, I have an issue.
>
> I need my (employee) users to be able to insert shows into the our  
> MySQL
> database and simultaneously upload an image file (and store the path  
> in the
> table).
>
> I have accomplished this with a product-based system (adding  
> products and
> uploading images of the product), and accomplished what I needed  
> because the
> product name was unique; I used the following statements:
>
> $prodName = $_POST['prodName'];
> $prodDesc = $_POST['prodDesc'];
> $prodPrice = $_POST['prodPrice'];
>
> $query2  = "INSERT INTO product (pID, pName) VALUES (NULL,  
> '$prodName');";
> $result2 = mysql_query($query2) or die(mysql_error());
>
> $query  = "SELECT pID FROM product WHERE pName = '$prodName';";
> $result = mysql_query($query) or die(mysql_error());
> $row = mysql_fetch_array($result) or die (mysql_error());
>
> $prodID = $row['pID'];
>
>
> I had to select the new product to get the product id to use in the  
> new
> unique image name.
>
> The problem I am facing now, is that with the shows that my users  
> add will
> have multitple show times; this means non-unique titles. In fact,  
> the only
> unique identifier is the show id. How can I insert something  
> (leaving the
> show_id field NULL so that it is auto-assigned the next ID number),  
> and then
> immediately select it?
>
> PHP doesn't seem to be able to immediately select something it has  
> just
> inserted, perhaps it needs time to process the database update.
>
> Here is the code I have now (which does not work):
>
> $query2  = "INSERT INTO afy_show (show_id, show_title, show_day_w,
> show_month, show_day_m, show_year, show_time, show_price,  
> show_description,
> show_comments_1, show_seats_reqd) VALUES (NULL, '{$show_title}',
> '{$show_day_w}', '{$show_month}', '{$show_day_m}', '{$show_year}',
> '{$show_time}', '{$show_price}', '{$show_description}',
> '{$show_comments_1}', '{$show_seats_reqd}');";
> $result2 = mysql_query($query2) or die(mysql_error());
>
> $query3 = "SELECT * FROM afy_show WHERE *show_id = '$id'*;";
> $result3 = mysql_query($query3) or die('Record cannot be located!' .
> mysql_error());
> $row3 = mysql_fetch_array($result3);
> $show_id = $row3['show_id'];
>
> How do I select the item I just inserted to obtain the ID number??

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[Floyd Resler-2]

Allen,
        Use mysql_insert_id().  This will return the id of the last record  
inserted.

Take care,
Floyd

On Oct 28, 2009, at 3:21 PM, Allen McCabe wrote:

> Hey everyone, I have an issue.
>
> I need my (employee) users to be able to insert shows into the our  
> MySQL
> database and simultaneously upload an image file (and store the path  
> in the
> table).
>
> I have accomplished this with a product-based system (adding  
> products and
> uploading images of the product), and accomplished what I needed  
> because the
> product name was unique; I used the following statements:
>
> $prodName = $_POST['prodName'];
> $prodDesc = $_POST['prodDesc'];
> $prodPrice = $_POST['prodPrice'];
>
> $query2  = "INSERT INTO product (pID, pName) VALUES (NULL,  
> '$prodName');";
> $result2 = mysql_query($query2) or die(mysql_error());
>
> $query  = "SELECT pID FROM product WHERE pName = '$prodName';";
> $result = mysql_query($query) or die(mysql_error());
> $row = mysql_fetch_array($result) or die (mysql_error());
>
> $prodID = $row['pID'];
>
>
> I had to select the new product to get the product id to use in the  
> new
> unique image name.
>
> The problem I am facing now, is that with the shows that my users  
> add will
> have multitple show times; this means non-unique titles. In fact,  
> the only
> unique identifier is the show id. How can I insert something  
> (leaving the
> show_id field NULL so that it is auto-assigned the next ID number),  
> and then
> immediately select it?
>
> PHP doesn't seem to be able to immediately select something it has  
> just
> inserted, perhaps it needs time to process the database update.
>
> Here is the code I have now (which does not work):
>
> $query2  = "INSERT INTO afy_show (show_id, show_title, show_day_w,
> show_month, show_day_m, show_year, show_time, show_price,  
> show_description,
> show_comments_1, show_seats_reqd) VALUES (NULL, '{$show_title}',
> '{$show_day_w}', '{$show_month}', '{$show_day_m}', '{$show_year}',
> '{$show_time}', '{$show_price}', '{$show_description}',
> '{$show_comments_1}', '{$show_seats_reqd}');";
> $result2 = mysql_query($query2) or die(mysql_error());
>
> $query3 = "SELECT * FROM afy_show WHERE *show_id = '$id'*;";
> $result3 = mysql_query($query3) or die('Record cannot be located!' .
> mysql_error());
> $row3 = mysql_fetch_array($result3);
> $show_id = $row3['show_id'];
>
> How do I select the item I just inserted to obtain the ID number??

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[Martin Scotta]

On Wed, Oct 28, 2009 at 4:21 PM, Allen McCabe <allenmccabe@...> wrote:

> Hey everyone, I have an issue.
>
> I need my (employee) users to be able to insert shows into the our MySQL
> database and simultaneously upload an image file (and store the path in the
> table).
>
> I have accomplished this with a product-based system (adding products and
> uploading images of the product), and accomplished what I needed because
> the
> product name was unique; I used the following statements:
>
> $prodName = $_POST['prodName'];
> $prodDesc = $_POST['prodDesc'];
> $prodPrice = $_POST['prodPrice'];
>
> $query2  = "INSERT INTO product (pID, pName) VALUES (NULL, '$prodName');";
> $result2 = mysql_query($query2) or die(mysql_error());
>
> $query  = "SELECT pID FROM product WHERE pName = '$prodName';";
> $result = mysql_query($query) or die(mysql_error());
> $row = mysql_fetch_array($result) or die (mysql_error());
>
> $prodID = $row['pID'];
>
>
> I had to select the new product to get the product id to use in the new
> unique image name.
>
> The problem I am facing now, is that with the shows that my users add will
> have multitple show times; this means non-unique titles. In fact, the only
> unique identifier is the show id. How can I insert something (leaving the
> show_id field NULL so that it is auto-assigned the next ID number), and
> then
> immediately select it?
>
> PHP doesn't seem to be able to immediately select something it has just
> inserted, perhaps it needs time to process the database update.
>
> Here is the code I have now (which does not work):
>
> $query2  = "INSERT INTO afy_show (show_id, show_title, show_day_w,
> show_month, show_day_m, show_year, show_time, show_price, show_description,
> show_comments_1, show_seats_reqd) VALUES (NULL, '{$show_title}',
> '{$show_day_w}', '{$show_month}', '{$show_day_m}', '{$show_year}',
> '{$show_time}', '{$show_price}', '{$show_description}',
> '{$show_comments_1}', '{$show_seats_reqd}');";
>  $result2 = mysql_query($query2) or die(mysql_error());
>
>  $query3 = "SELECT * FROM afy_show WHERE *show_id = '$id'*;";
>  $result3 = mysql_query($query3) or die('Record cannot be located!' .
> mysql_error());
>  $row3 = mysql_fetch_array($result3);
>  $show_id = $row3['show_id'];
>
> How do I select the item I just inserted to obtain the ID number??
>

mysql_insert_id <http://ar.php.net/manual/en/function.mysql-insert-id.php>
mysqli->insert_id <http://ar.php.net/manual/en/mysqli.insert-id.php>


--
Martin Scotta

[Allen McCabe]

You all are great! Thank you so much.



On Wed, Oct 28, 2009 at 12:27 PM, Martin Scotta <martinscotta@...>wrote:

>
>
> On Wed, Oct 28, 2009 at 4:21 PM, Allen McCabe <allenmccabe@...>wrote:
>
>> Hey everyone, I have an issue.
>>
>> I need my (employee) users to be able to insert shows into the our MySQL
>> database and simultaneously upload an image file (and store the path in
>> the
>> table).
>>
>> I have accomplished this with a product-based system (adding products and
>> uploading images of the product), and accomplished what I needed because
>> the
>> product name was unique; I used the following statements:
>>
>> $prodName = $_POST['prodName'];
>> $prodDesc = $_POST['prodDesc'];
>> $prodPrice = $_POST['prodPrice'];
>>
>> $query2  = "INSERT INTO product (pID, pName) VALUES (NULL, '$prodName');";
>> $result2 = mysql_query($query2) or die(mysql_error());
>>
>> $query  = "SELECT pID FROM product WHERE pName = '$prodName';";
>> $result = mysql_query($query) or die(mysql_error());
>> $row = mysql_fetch_array($result) or die (mysql_error());
>>
>> $prodID = $row['pID'];
>>
>>
>> I had to select the new product to get the product id to use in the new
>> unique image name.
>>
>> The problem I am facing now, is that with the shows that my users add will
>> have multitple show times; this means non-unique titles. In fact, the only
>> unique identifier is the show id. How can I insert something (leaving the
>> show_id field NULL so that it is auto-assigned the next ID number), and
>> then
>> immediately select it?
>>
>> PHP doesn't seem to be able to immediately select something it has just
>> inserted, perhaps it needs time to process the database update.
>>
>> Here is the code I have now (which does not work):
>>
>> $query2  = "INSERT INTO afy_show (show_id, show_title, show_day_w,
>> show_month, show_day_m, show_year, show_time, show_price,
>> show_description,
>> show_comments_1, show_seats_reqd) VALUES (NULL, '{$show_title}',
>> '{$show_day_w}', '{$show_month}', '{$show_day_m}', '{$show_year}',
>> '{$show_time}', '{$show_price}', '{$show_description}',
>> '{$show_comments_1}', '{$show_seats_reqd}');";
>>  $result2 = mysql_query($query2) or die(mysql_error());
>>
>>  $query3 = "SELECT * FROM afy_show WHERE *show_id = '$id'*;";
>>  $result3 = mysql_query($query3) or die('Record cannot be located!' .
>> mysql_error());
>>  $row3 = mysql_fetch_array($result3);
>>  $show_id = $row3['show_id'];
>>
>> How do I select the item I just inserted to obtain the ID number??
>>
>
> mysql_insert_id <http://ar.php.net/manual/en/function.mysql-insert-id.php>
> mysqli->insert_id <http://ar.php.net/manual/en/mysqli.insert-id.php>
>
>
> --
> Martin Scotta
>