Post Query into Form

[cassy01]

Hi There,

 

i am trying to get the following query into a form;

$query = "SELECT * FROM products WHERE itemID = '$_POST[itemID]'";

 

i am then trying to put each field in that row into a form so the user can edit it and submit it back to the database, i have tried the following;

 

$result=mysql_query($query);

$row = mysql_fetch_array($result)

$itemID =  $result['itemID']

$itemName =  $result['itemName']

 

<input name="itemID" type="text" id="itemID" value="$itemID" />

 

please help

[monkeybidz]

Try this:

<?php
$query = "SELECT * FROM products WHERE itemID = ".$_POST[itemID];

?>

[cassy01]

thats what ive already got :S

 

i can get the data from the database using the above query but cannot get each field to be inserted into the form.

[AndyB]

$itemID =  $result['itemID']

$itemName =  $result['itemName']

 

should be

 

$itemID =  $row['itemID']

$itemName =  $row['itemName']

 

$row is the array you abstracted $result into!

[cassy01]

now i have changed it to row it comes up with;

 

Parse error: syntax error, unexpected T_VARIABLE

 

 

[AndyB]

The actual code lines I gave as example would need end with ;

 

Fx:

 

$itemID =  $row['itemID'];
$itemName =  $row['itemName'];

 

Can't fix what we can't see 

 

If you still have problems, post the relevant parts of the form processor code

[cassy01]

right ive got the correct code at the top;

 

<?php $query = "SELECT * FROM products WHERE itemID = '$_POST[itemID]'";

$result=mysql_query($query);

$row = mysql_fetch_array($result);

?>

 

now how can i get the row of data to be entered as values to the form? i have at the moment;

 

<input name="itemName" type="text" id="itemName" value="<?php $row['itemName'];?>" />

[monkeybidz]

thats what ive already got :S

 

i can get the data from the database using the above query but cannot get each field to be inserted into the form.

 

Look good at how I posted the code in the query at the end with different quotes. Makes a difference. If that does not work, try:

<?php
$query = "SELECT * FROM products WHERE itemID = ".$_POST['itemID'];

?>

[cassy01]

changed code to:

 

$query = "SELECT * FROM products WHERE itemID = ".$_POST['itemID'];

$result=mysql_query($query);

$row = mysql_fetch_array($result);

 

Now trying to put that row of data into a form as values;

 

<input name="itemName" type="text" id="itemName" value="<?php $row['itemName'];?>"  />

[AndyB]

<input name="itemName" type="text" id="itemName" value="<?php $row['itemName'];?>"  />

 

should be:

 

<input name="itemName" type="text" id="itemName" value="<?php echo $row['itemName'];?>"  />

[cassy01]

thanks guys works perfectly