CODEWITHSUNDEEP
phphtml
charlie_cat
charlie_cat

Reputation: 1850

My form "post" is not working

here is only part of my code, the action does not well ...action. when i view in the google javascript tools i have an id in the action line when i hit the submit button, nothing happens, its suppose to go to php/form.php with ID 28 in this instance

 ......
 if (mysqli_num_rows($result))
 {
    while ($row = mysqli_fetch_assoc($result)) 
   {  
?>
<form name="form-list" method="post" action='php/form.php?id=<?php echo $row["id"]; ?>' >
 <table> 
   <tr>
     <td valign="top">
        <label for="surname">Surname</label>
     </td>
     <td valign="top"> 
       <?php echo $row["title"]; ?> 
     </td>                           
   </tr>                                
   <tr>
     <td valign="top"> 
       <label for="name">Id</label> 
     </td>                   
     <td valign="top"> 
      <input type="text" value='<?php echo $row["id"]; ?>' name="name" maxlength="50" size="30"> 
     </td>        
   </tr>     
   <tr>
     <td valign="top">            
       <input type="button" value="Submit" />   
     </td>
   </tr>        
 <?php } } ?>   
</table>    
</form>  .....

enter image description here

Upvotes: 0

Views: 77

Answers (1)

Xyaren
Xyaren

Reputation: 953

<input type="submit" value="Submit" />

The input type is not button it is submit. The value attribute just changes the written text on the button but does not change it's behavior.

Note to the other answers:

Using POST is fine if he handles the request in the right way. In the given code example everything except the submit button works and is fine.

In this case the id is transmitted as a GET and all other values are submitted via POST. You can mix them.

Later you do have to use $_GET["id"] and $_POST["name"].. or you can even use $_REQUEST["..."] for all of them.

Upvotes: 4

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