Undefined index: Project_Name

Question

I'm trying to count all my project in project table Project_Name is a column_name

Here is the my code I have tried:

<?php
   $sql = "SELECT COUNT(*) FROM project";
   $result = $connection->query($sql);  
   if ($result->num_rows > 0) {
      $row = $result->fetch_assoc();    
      $project_count = $row['Project_Name'];
   } 
   else {
      echo "0 results"; 
   } 
?>

And what output do you get? Note that `Project_Name` is not present in your `SELECT` query so it won't be in the results either. — Val, May 14 '15 at 04:26
possible duplicate of [PHP Undefined Index](http://stackoverflow.com/questions/4842759/php-undefined-index) — Sougata Bose, May 14 '15 at 04:26

Val · Accepted Answer · 2015-05-14T04:33:31.450

0

Try this

$sql = "SELECT COUNT(*) FROM project";
$result = $connection->query($sql); 
if ($result->num_rows > 0) {
    $row =$result->fetch_array();
    $project_count = $row[0];
}

edited May 14 '15 at 04:33

answered May 14 '15 at 04:27

Val

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`mysql_fetch_array` you can't mix `mysql_` functions with anything other than their own. That should read as `$result->fetch_array()` since they're using an object oriented method to connect with. – Funk Forty Niner May 14 '15 at 04:32
`num_rows` doesn't make sense, it will always yield a row, just get the count – Kevin May 14 '15 at 04:35
thanks it works but i found way out by echo $project_count[0]; with an array but it doesnt make sense but this one should work thanks a lot! – Derek Anas May 14 '15 at 05:32

score 0 · Answer 2 · edited May 14 '15 at 04:53

0

The query is returning only the count. Use another query like: SELECT * FROM project; for fetching the columns values in each row. With this, you will be able to do $row['Project_Name'];

edited May 14 '15 at 04:53

Val

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answered May 14 '15 at 04:27

shruti1810

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Undefined index: Project_Name

2 Answers2