cefalufr
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Posts posted by cefalufr
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It Keeps Failing on insertion, someone point of my error? DB information is above this code btw!
[code]$fname = $_REQUEST['fname'];
$lname = $_REQUEST['lname'];
$state = $_REQUEST['state'];
$state = $_REQUEST['city'];
$zip = $_REQUEST['zip'];
$userid = $_REQUEST['userid'];
$query= 'INSERT INTO shipping_multiple (fname, lname, state, city, zip, userid) VALUES ("'.$fname.'","'.$lname.'","'.$state.'","'.$city.'","'.$zip.'","'.$userid.'")';
mysql_query($query) or die('Failed Insertion For New User');[/code]
[code]
$fname = $_REQUEST['fname'];
$lname = $_REQUEST['lname'];
$state = $_REQUEST['state'];
$city = $_REQUEST['city'];
$zip = $_REQUEST['zip'];
$userid = $_REQUEST['shipping_multiple'];
echo $fname;
echo $lname;
echo $state;
echo $city;
echo $zip;
echo $userid;
$query= 'INSERT INTO shipping_multiple (fname, lname, state, city, zip, userid) VALUES ("'.$fname.'","'.$lname.'","'.$state.'","'.$city.'","'.$zip.'","'.$userid.'")';
mysql_query($query) or die('Failed Insertion For New User');[/code]
It Echo's Back the Forms Variables perfectly but fails on insertion
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I tried using Dreamweaver to handle PHP Database Coding for about 2 days. Dont. Its horrible. Its better off learning how to do it.
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[code]<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = 'this';
$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql');
$dbname = 'oscommerce_feedyoursoul';
mysql_select_db($dbname) or die('Failed to Select database');
?>
<?php
$connect = 'SELECT * FROM `shipping_multiple`';
mysql_query($connect) or die('failed query: ' . $connect . '<br>' . mysql_error());
$num_rows = mysql_num_rows($connect);
$i = '0'
?>
<SELECT name="product" size="5" multiple>
<OPTION SELECTED>
<?php
while($i < $num_rows)
{
$firstname = mysql_result("fname");
$lastname = mysql_result("lname");
echo $firstname . $lastname;
}
?>[/code] -
[!--quoteo(post=371069:date=May 3 2006, 04:55 PM:name=kenrbnsn)--][div class=\'quotetop\']QUOTE(kenrbnsn @ May 3 2006, 04:55 PM) [snapback]371069[/snapback][/div][div class=\'quotemain\'][!--quotec--]
Modify the "or die" clause on the mysql_query statement to output the mysql_error() and the query.
[code]<?php
$connect = 'SELECT fname lname FROM shipping_multiple';
mysql_query($connect) or die('failed query: ' . $connect . '<br>' . mysql_error());
?>[/code]
I believe you're missing a comma between the field names in your query.
Ken
[/quote]
Still Same Error, This is so annoying.
[code]<?php
$connect = 'SELECT `fname` ,`lname` FROM shipping_multiple';
mysql_query($connect) or die('failed query: ' . $connect . '<br>' . mysql_error());
$num_rows = mysql_num_rows($connect);
$i = '0'
?>[/code] -
I keep recieving this Error = [!--quoteo--][div class=\'quotetop\']QUOTE[/div][div class=\'quotemain\'][!--quotec--]Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource ....[/quote]
Help? Also My DB Connection is above this code. The query returns no errors, I has to do with the num_rows.
[code]<?php
$connect = 'SELECT fname lname FROM shipping_multiple';
mysql_query($connect) or die('failed query');
$num_rows = mysql_num_rows($connect);
$i = '0'
?>
<SELECT name="product" size="5" multiple>
<OPTION SELECTED>
<?php
while($i < $num_rows)
{
$firstname = mysql_result("fname");
$lastname = mysql_result("lname");
echo $firstname . $lastname;
}
?>
</OPTION></SELECT></FORM>[/code]
Failing on insertion
in PHP Coding Help
Posted
echo the query and then copy/paste it directly into phpmyadmin see if it inserts the row without issues.
also, changing this:
...or die('Failed Insertion For New User');
to this:
... or die(mysql_error());
will give you better help for what's wrong.
[/quote]
Thks. Never thought of checking. Wrong column name for userid!