php display results from mysql database

Question

Not done any php database work for years, fairly simple question I'm sure for anyone with experience, below is my code it should take all the results from the database and display these to a table with links:

<?php
include('session.php');
?>
<html">
<head>
<title>Welcome </title>
</head> 
<body>
  <h1>Welcome <?php echo $login_session; ?></h1> 

  <form method="post" action="upload.php" enctype="multipart/form-data">
<p>File:</p>
<input type="file" name="Filename"> 
<p>Description:</p>
<textarea rows="10" cols="35" name="Description"></textarea>
<br/>
<input TYPE="submit" name="upload" value="Submit"/>
</form>

<hr/>
<p>Uploaded Files</p>

<?php
//include('config.php');
$query1=mysqli_query($db,"SELECT filepath,filename,description FROM `filedetails`");
echo "<table><tr><td>filepath</td><td>filename</td><td></td><td></td>";
while($query2=mysql_fetch_array($query1))
{
echo "<tr><td>".$query2['filepath']."</td>";
echo "<td>".$query2['filename']."</td>";
echo "<td><a href='edit.php?id=".$query2['id']."'>Edit</a></td>";
echo "<td><a href='delete.php?id=".$query2['id']."'>x</a></td><tr>";
}
?>
<h2><a href = "logout.php">Sign Out</a></h2>
</body>
</html>

So the line it's failing on is

while($query2=mysql_fetch_array($query1))

The error message is:

Warning: mysql_fetch_array() expects parameter 1 to be resource, object given in C:\xampp\htdocs\welcome.php on line 27

You're mixing both MySQLi (mysqli_query) and MySQL (mysql_fetch_array) syntax. MySQL functions are deprecated in PHP 5.5, and removed in PHP 7, so please don't use it. — Obsidian Age
– Obsidian Age, Commented Apr 4, 2017 at 22:37

Nick Coad · Accepted Answer · 2017-04-04 22:42:11Z

Instead of mysql_fetch_array you need to use mysqli_fetch_array.

<?php
    include('session.php');
?>
<html">
    <head>
        <title>Welcome </title>
    </head> 
    <body>
        <h1>Welcome <?php echo $login_session; ?></h1> 

        <form method="post" action="upload.php" enctype="multipart/form-data">
            <p>File:</p>
            <input type="file" name="Filename"> 
            <p>Description:</p>
            <textarea rows="10" cols="35" name="Description"></textarea>
            <br/>
            <input TYPE="submit" name="upload" value="Submit"/>
        </form>

        <hr/>
        <p>Uploaded Files</p>

        <?php
            //include('config.php');
            $query1=mysqli_query($db,"SELECT filepath,filename,description FROM `filedetails`");
            echo "<table><tr><td>filepath</td><td>filename</td><td></td><td></td>";
            while($row=mysqli_fetch_array($query1))
            {
                echo "<tr><td>".$row['filepath']."</td>";
                echo "<td>".$row['filename']."</td>";
                echo "<td><a href='edit.php?id=".$row['id']."'>Edit</a></td>";
                echo "<td><a href='delete.php?id=".$row['id']."'>x</a></td><tr>";
            }
        ?>
        <h2><a href = "logout.php">Sign Out</a></h2>
    </body>
</html>

Note: I also changed your variable name to a more sensible one.

TypedSource · Accepted Answer · 2017-04-04 22:44:04Z

1

your resource is mysqli and you try to handle with mysql_fetch_array. try to use mysqli_fetch_array. i think this will eliminate your problem

$resource = mysqli_query($connection, $queryString);
if(!mysqli_errno($connection)) {
  while($resultArray = mysqli_fetch_array($resource)){
    print_r($resultArray); // pseudo output for your table
  }
} else {
  echo "error\n"
  print_r(mysqli_error($connection))
}

And never forget to check for errors on the query

answered Apr 4, 2017 at 22:44

TypedSource

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Comments

Brijesh_yadav · Accepted Answer · 2017-04-05 12:37:44Z

1

The error lies in the mismatch of mysql and mysqli connections. As you are using mysqli_query, you should use the mysqli. Replacing the following line with error line should do the work:

mysqli_fetch_array($db,$query1);

answered Apr 5, 2017 at 12:37

Brijesh_yadav

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