Solved

PHP Form/Mysql Issues

Posted on 2011-03-08

1 solution

195 Views

Last Modified: 2013-12-25

Okay,
I've taken some of your advice, example codes, read some URLs & have build the following but it is not pulling the data from my mysql database so I can see it to edit it. Can soomeone take a look at this to see where my mistake is? I've also attached a photo of the fields I'm trying to edit to help you better understand what I am doing. Thanks

<?php

$con = mysql_connect("localhost","uname","password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("mydb", $con);

$result = mysql_query("SELECT * FROM mytable ");

$i=0;


while ($i < $num) {
$due_ge=mysql_result($result,$i,"due_ge");
$due_rk=mysql_result($result,$i,"due_rk");
$due_ny=mysql_result($result,$i,"due_ny");

?>

<form action="updated.php">
<input type="hidden" name="id" value="<? echo "$id"; ?>">
DATE1: <input type="date" name="due_ge" value="<? echo "$due_ge"?>"><br>
DATE2: <input type="date" name="due_rk" value="<? echo "$due_rk"?>"><br>
DATE3: <input type="date" name="due_ny" value="<? echo "$due_ny"?>"><br>
<input type="Submit" value="Update">
</form>

<?php
++$i;
} 
?>

Open in new window

editeditedit.JPG

Comment

Question by:wantabe2

[X]

Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

Help others & share knowledge
Earn cash & points
Learn & ask questions

Join The Community

3 Comments

LVL 27

Expert Comment

by:Lukasz Chmielewski

ID: 350713972011-03-08

Use this at the top to see what's wrong

$con = mysql_connect("localhost","uname","password") or die(mysql_error());
mysql_select_db("mydb", $con) or die(mysql_error());;

$result = mysql_query("SELECT * FROM mytable ") or die(mysql_error());;

LVL 27

Accepted Solution

by:Lukasz Chmielewski

Lukasz Chmielewski earned 500 total points

ID: 350714312011-03-08

Aso, there is no $num variable

<?php
$con = mysql_connect("localhost","uname","password") or die(mysql_error());
mysql_select_db("mydb", $con) or die(mysql_error());;
$result = mysql_query("SELECT * FROM mytable ") or die(mysql_error());;
$num = mysql_num_rows($result);
$i=0;

while ($i < $num) {
$due_ge=mysql_result($result,$i,"due_ge");
$due_rk=mysql_result($result,$i,"due_rk");
$due_ny=mysql_result($result,$i,"due_ny");

?>

<form action="updated.php">
<input type="hidden" name="id" value="<? echo "$id"; ?>">
DATE1: <input type="date" name="due_ge" value="<? echo "$due_ge"?>"><br>
DATE2: <input type="date" name="due_rk" value="<? echo "$due_rk"?>"><br>
DATE3: <input type="date" name="due_ny" value="<? echo "$due_ny"?>"><br>
<input type="Submit" value="Update">
</form>

<?php
++$i;
}

Open in new window

LVL 27

Expert Comment

by:Lukasz Chmielewski

ID: 350714602011-03-08

I've noticed that there is no $id declared and for each record there is a form generated - is that correct thing to do ?

add
error_reporting(E_ALL);
to the top of the script and this will tell you more.

Featured Post

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Improved Form Tokens to Guard Against CSRF and Screen Scrapers

Article by: Ray

Nothing in an HTTP request can be trusted, including HTTP headers and form data. A form token is a tool that can be used to guard against request forgeries (CSRF). This article shows an improved approach to form tokens, making it more difficult to…