PHP Fatal error: Call to undefined function test_input() in C:\wamp\www\web\new9.php on line 11

Question

The following form shows an error on clicking the submit button if we enter the name only and submit.The error shown is

Fatal error: Call to undefined function test_input() in C:\wamp\www\web\new9.php on line 11

Can anyone find the problem in the below code.

      <?php
// define variables and set to empty values
$name1Err = $email1Err  =  "";
$name1 = $email1 =  "";

if ($_SERVER["REQUEST_METHOD"] == "POST")
{
 if (empty($_POST["input_1"]))
    {$name1Err = "Your name is required. Just the first will do. ";}
  else
    {$name1 = test_input($_POST["input_1"]);
     if (!preg_match("/^[a-zA-Z ]*$/",$name1))
       {       $name1Err = "Only letters and white space allowed"; }
    }

 if (empty($_POST["input_12"]))
    {$email1= "";}
  else
    {$email1 = test_input($_POST["input_12"]);
    // check if e-mail address syntax is valid
     if (!preg_match("/([\w\-]+\@[\w\-]+\.[\w\-]+)/",$email1))
       {  $emailErr = "Invalid email format"; }
    }




function test_input($data)
{
   $data = trim($data);
   $data = stripslashes($data);
   $data = htmlspecialchars($data);
   return $data;
}
}
?>
        <form method='post' enctype='multipart/form-data' action=''>
                name<input name='input_1' type='text' value='<?php echo $name1;?>' tabindex='1' />
                <div class="validation_message"><?php echo $name1Err ?></div>

                email<input name='input_12'  type='email' value='<?php echo $email1;?>'/><br/>
                <textarea name='input_5' tabindex='9'   rows='10' cols='50'></textarea>

            <input type='submit' value='Submit' tabindex='25' />

    </form>


<?php
echo "<h2>Your Input:</h2>";
echo "Name:".$name1;
echo "<br>";
echo "Email:".$email1;
echo "<br>";
?>

just a side note, your function could be optimised, $data = htmlentities(trim($data)); return $data; — Pwner
– Pwner, Commented Jan 15, 2014 at 14:04

Shankar Narayana Damodaran · Accepted Answer · 2014-01-15 14:01:41Z

8

Move your function out of your conditional if statement

It should be like this...

<?php
// define variables and set to empty values
$name1Err = $email1Err  =  "";
$name1 = $email1 =  "";

// Moved here
function test_input($data)
{
   $data = trim($data);
   $data = stripslashes($data);
   $data = htmlspecialchars($data);
   return $data;
}

if ($_SERVER["REQUEST_METHOD"] == "POST")
{
// .... your remaining code .......... !

From the PHP Docs...

When a function is defined in a conditional manner ... Its definition must be processed prior to being called.

Source

answered Jan 15, 2014 at 14:01

Shankar Narayana Damodaran

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1 Comment

Niet the Dark Absol Over a year ago

You may want to explain the reason for needing to do this.

Niet the Dark Absol · Accepted Answer · 2014-01-15 14:05:12Z

In general, functions are parsed first and can therefore be used in any order.

echo foo();
function foo() {return "bar";}

The above works fine.

However, unlike some languages like JavaScript, PHP allows you to conditionally define functions. You might do something like this:

if( $something) {
    function foo() {echo "bar";}
}
else {
    function foo() {echo "fish";}
}
foo();

It's a bad thing to do (personally I'd prefer anonymous functions, or putting the conditional inside the function), but it's allowed.

However, doing this means that the functions can no longer be grabbed. They MUST be defined before they can be used. Going back to our first example:

if( true) {
    echo foo();
    function foo() {return "bar";}
}

This will fail.

Yotam Omer · Accepted Answer · 2015-02-01 09:26:08Z

1

try

<form method='post' enctype='multipart/form-data' action='<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>'>

source:w3schools

edited Feb 1, 2015 at 9:26

Yotam Omer

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answered Oct 19, 2014 at 8:39

Arnaud

111 bronze badge

Comments

yassine afaila · Accepted Answer · 2023-05-30 10:40:18Z

0

you need to define the function first then you can use it , i just found this but when i did with javascript it work totally fine

answered May 30, 2023 at 10:40

yassine afaila

214 bronze badges

1 Comment

DSDmark Over a year ago

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