未定义的索引错误PHP

Question

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!

Answer 1

Try:

<?php

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

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

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

?>

Answer 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'] : '';

Answer 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'];

}

Answer 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
         }

Answer 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.

Answer 6

嘿这发生了，因为你试图在赋值之前显示价值你只需填写值并提交表格就会显示正确的输出或者你可以写下你的PHP代码下面的表格标签它会运行没有任何错误

Answer 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

Answer 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

Answer 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 ie if you start with form-data in enctype, so you should start with multipart:

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