Undefined index error PHP

问题:

I'm new in PHP and I'm getting this error:

Notice: Undefined index: productid in /var/www/test/modifyform.php on line 32

Notice: Undefined index: name in /var/www/test/modifyform.php on line 33

Notice: Undefined index: price in /var/www/test/modifyform.php on line 34

Notice: Undefined index: description in /var/www/test/modifyform.php on line 35

I couldn't find any solution online, so maybe someone can help me.

Here is the code:

<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="POST">
   <input type="hidden" name="rowID" value="<?php echo $rowID;?>">

   <p>
      Product ID:<br />
      <input type="text" name="productid" size="8" maxlength="8" value="<?php echo $productid;?>" />
   </p>

   <p>
      Name:<br />
      <input type="text" name="name" size="25" maxlength="25" value="<?php echo $name;?>" />
   </p>

   <p>
      Price:<br />
      <input type="text" name="price" size="6" maxlength="6" value="<?php echo $price;?>" />
   </p>

   <p>
      Description:<br />
      <textarea name="description" rows="5" cols="30">
      <?php echo $description;?></textarea>
   </p>

   <p>
      <input type="submit" name="submit" value="Submit!" />
   </p>
   </form>
   <?php
   if (isset($_POST['submit'])) {
      $rowID = $_POST['rowID'];
      $productid = $_POST['productid']; //this is line 32 and so on...
      $name = $_POST['name'];
      $price = $_POST['price'];
      $description = $_POST['description'];

}

What I do after that (or at least I'm trying) is to update a table in MySQL. I really can't understand why $rowID is defined while the other variables aren't.

Thank you for taking your time to answer me. Cheers!

回答1:

Try:

<?php

if (isset($_POST['name'])) {
    $name = $_POST['name'];
}

if (isset($_POST['price'])) {
    $price = $_POST['price'];
}

if (isset($_POST['description'])) {
    $description = $_POST['description'];
}

?>

回答2:

Apparently the index 'productid' is missing from your html form. Inspect your html inputs first. eg <input type="text" name="productid" value=""> But this will handle the current error PHP is raising.

  $rowID = isset($_POST['rowID']) ? $_POST['rowID'] : '';
  $productid = isset($_POST['productid']) ? $_POST['productid'] : '';
  $name = isset($_POST['name']) ? $_POST['name'] : '';
  $price = isset($_POST['price']) ? $_POST['price'] : '';
  $description = isset($_POST['description']) ? $_POST['description'] : '';

回答3:

TRY

<?php

  $rowID=$productid=$name=$price=$description="";  

   if (isset($_POST['submit'])) {
      $rowID = $_POST['rowID'];
      $productid = $_POST['productid']; //this is line 32 and so on...
      $name = $_POST['name'];
      $price = $_POST['price'];
      $description = $_POST['description'];

}

回答4:

This is happening because your PHP code is getting executed before the form gets posted.

To avoid this wrap your PHP code in following if statement and it will handle the rest no need to set if statements for each variables

       if(isset($_POST) && array_key_exists('name_of_your_submit_input',$_POST))
        {
             //process PHP Code
        }
        else
        {
             //do nothing
         }

回答5:

There should be the problem, when you generate the <form>. I bet the variables $name, $price are NULL or empty string when you echo them into the value of the <input> field. Empty input fields are not sent by the browser, so $_POST will not have their keys.

Anyway, you can check that with isset().

Test variables with the following:

if(isset($_POST['key'])) ? $variable=$_POST['key'] : $variable=NULL

You better set it to NULL, because

NULL value represents a variable with no value.

回答6:

Hey this is happening because u r trying to display value before assignnig it U just fill in the values and submit form it will display correct output Or u can write ur php code below form tags It ll run without any errors

回答7:

If you are using wamp server , then i recommend you to use xampp server . you . i get this error in less than i minute but i resolved this by using (isset) function . and i get no error . and after that i remove (isset) function and i don,t see any error.

by the way i am using xampp server

回答8:

this error occurred sometime method attribute ( valid passing method ) Error option : method="get" but called by $Fname = $_POST["name"]; or

       method="post" but  called by  $Fname = $_GET["name"];

More info visit http://www.doordie.co.in/index.php

回答9:

To remove this error, in your html form you should do the following in enctype:

<form  enctype="multipart/form-data">

The following down is the cause of that error i.e if you start with form-data in enctype, so you should start with multipart:

<form enctype="form-data/multipart">