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<?php if (isset($_GET['hash'])&&!empty($_GET['hash'])){ $hash = $_GET['hash']; $message_query = "SELECT from_id, message FROM message WHERE hash='$hash'"; $run_messages = mysqli_query($con,$message_query); while($row_messages = mysqli_fetch_array($con,$run_messages)){ $form_id = $row_messages['from_id']; $message = $row_messages['message']; $user_query = "SELECT username FROM admins WHERE id='$from_id'"; $run_user = mysqli_fetch_array($con,$user_query); $from_username = $run_user['username']; echo "<p><strong>$from_username</strong></p></br>"; } }else{ header('Location: messages.php'); } ?>

I'm getting this error message:

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, object given on line 6

Here's line 6 & as you can see I have already included the $con which is my database connection.

while($row_messages = mysqli_fetch_array($con,$run_messages)){

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  • printf("Error: %s\n", $mysqli->error); Commented Dec 18, 2015 at 11:57
  • If not solved, then Please post your database connection code. Commented Dec 18, 2015 at 12:19
  • Plus you having typo in second query $form_id !=$from_id it would be SELECT username FROM admins WHERE id=$form_id Commented Dec 18, 2015 at 12:19

2 Answers 2

Reset to default
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mysqli_fetch_array

mysqli_fetch_array ( mysqli_result $result [, int $resulttype = MYSQLI_BOTH ] )

First parameter is Specifies a result set identifier returned by mysqli_query() and second parameter is result type

Remove connection as first parameter

It would be

$row_messages = mysqli_query($run_messages,MYSQLI_ASSOC);

Your code is open for sql injection better use bind statement

http://php.net/manual/en/mysqli-stmt.bind-param.php

To check error in your connection and query use

/* check connection */ if ($mysqli->connect_errno) { printf("Connect failed: %s\n", $mysqli->connect_error); exit(); } if (!$mysqli->query("SET a=1")) { printf("Errormessage: %s\n", $mysqli->error); }

http://php.net/manual/en/mysqli.error.php

In second query you use

$user_query = "SELECT username FROM admins WHERE id='$from_id'";

TYPO here

$form_id !=$from_id

Change this to

$user_query = "SELECT username FROM admins WHERE id=' $form_id'";

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3 Comments

The 2 errors appear: : mysqli_query() expects parameter 1 to be mysqli, string given on line 5 & mysqli_fetch_array() expects parameter 1 to be mysqli_result, object given on line 6
Have your problem solved using $row_messages = mysqli_query($run_messages,MYSQLI_ASSOC); ??
You have typo in second query $form_id !=$from_id
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<?php if (isset($_GET['hash'])&&!empty($_GET['hash'])){ $hash = mysqli_escape_string($con, $_GET['hash']); $message_query = "SELECT from_id, message FROM message WHERE hash='$hash'"; $run_messages = mysqli_query($con,$message_query); while($row_messages = mysqli_fetch_array($run_messages, MYSQLI_ASSOC)){ $from_id = $row_messages['from_id']; $message = $row_messages['message']; $user_query = "SELECT username FROM admins WHERE id='$from_id'"; $query_run = mysqli_query($con, $user_query); $run_user = mysqli_fetch_array($query_run, MYSQLI_ASSOC); $from_username = $run_user['username']; echo "<p><strong>$from_username</strong></p></br>"; } }else{ header('Location: messages.php'); } ?>

Essentially mysqli_fetch_array has one required parameter which is the result of a query and an optional query of the result type. $run_messages is the result of the first query and will be used for the execution of mysql_fetch_array and MYSQLI_ASSOC is the optional type that you will be using so you can access the values in $row_messages like you do.

Read mysqli_fetch_array for more information.

Also, please remember to escape user valued with mysqli_escape_string before allowing the data to be queried into the database. This reduces the chance of SQLi.

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2 Comments

Correct typo of second query in your code too $form_id!=$from_id
I was just going off what was given in the OP. I have changed it though.

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