mysqli_fetch_assoc() not working

Question

Could someone debug this and tell me why is is not working?

<?php
include("C:\Wamp\www\system\db\connect.php");
$term = mysqli_real_escape_string($con, $_GET['q']);
echo "results for \"".$term."\".<br>";
$sql = "SELECT * FROM `search` WHERE Keywords='%{$term}%' LIMIT 10";
$result = mysqli_query($con, $sql) or die("<p color=\"#f00\">Could not query database.</p>");
while($row = mysqli_fetch_assoc($result) or die("<p color=\"#f00\">Could not fetch assoc array in database.</p>")) {
    echo $row['Title'];
}
echo json_encode($row['Title']);

mysqli_close($con);
?>

It stops working at the mysqli_fetch_assoc function.

What exactly is not working ? you connection to the database ? your results are not the ones expected ? you have a php error ? — 0x1gene, Mar 23 '15 at 09:57
WHERE Keywords LIKE , not '=', unless you expect to get the string with percent characters on both ends — n-dru, Mar 23 '15 at 09:58
@0x1gene I can connect to the database fine. As it says is the title, the problem is with the `mysqli_fetch_assoc` function. The function returns the `die` string. — Krii, Mar 23 '15 at 09:58
Try printing `mysqli_num_rows($result)` before while loop to check if rows exist. — Apul Gupta, Mar 23 '15 at 10:01
Also, instead of printing custom error message, use `or die(mysqli_error())`. this will help you in identification of error. — Apul Gupta, Mar 23 '15 at 10:02
As you are using mysqli, that should be `or die(mysqli_error($con))` — mhall, Mar 23 '15 at 10:04
@mhall `Warning: mysql_error() expects parameter 1 to be resource, object given in` . — Krii, Mar 23 '15 at 10:06

score 0 · Answer 1 · answered Mar 23 '15 at 10:08

As mentioned in comments, you should print actual mysql error message instead of customized messages, it will help you in debugging the error,

Here are some suggestions to fix your bug,

Your code should be:

<?php
include("C:\Wamp\www\system\db\connect.php"); //<-- You should give relative path instead of this one
$term = mysqli_real_escape_string($con, $_GET['q']);
echo "results for \"".$term."\".<br>";
$sql = "SELECT * FROM `search` WHERE Keywords like '%{$term}%' LIMIT 10";
$result = mysqli_query($con, $sql) or die(mysqli_error($con)); //<-- show mysql error instead of custom one
while($row = mysqli_fetch_assoc($result) ) {
    echo $row['Title'];
}
echo json_encode($row['Title']);

mysqli_close($con);
?>

I used the path because using a relative one returns `failed to open stream: No such file or directory` but I have been to the directory in the browser. — Krii, Mar 23 '15 at 10:12

score 0 · Answer 2 · answered Mar 23 '15 at 10:13

Replace these lines:

$sql = "SELECT * FROM `search` WHERE Keywords='%{$term}%' LIMIT 10";
$result = mysqli_query($con, $sql) or die("<p color=\"#f00\">Could not query database.</p>");
while($row = mysqli_fetch_assoc($result) or die("<p color=\"#f00\">Could not fetch assoc array in database.</p>")) {
   echo $row['Title'];
}

with these:

$sql = "SELECT * FROM `search` WHERE Keywords LIKE '%{$term}%' LIMIT 10";
$result = mysqli_query($con, $sql) or die("<p color=\"#f00\">Could not query database.</p>");
while($row = mysqli_fetch_assoc($result) ) {
   echo $row['Title'];
}

Returns 'null' and if `die` is added, returns `die` string. – Krii Mar 23 '15 at 10:16 — Krii, Mar 23 '15 at 10:16

score 0 · Answer 3 · answered Jun 13 '18 at 11:44

I had the same issue on a simple table with one value, I added the die error messages and it was still showing nothing for me, try to echo one specific row of your table, for me it did show a result, so I figured my problem was in the encode as json. So i used the solution found in json encoding returning empty string As for the echo of your specific row, I tried this:

$result = mysqli_query($conn, $query) or die(mysqli_error($conn));

if($result->num_rows > 0){
     while($row= mysqli_fetch_assoc($result)){
        echo($row['name_ofRow']);
   }
}

I hope this will help some how.

mysqli_fetch_assoc() not working

3 Answers3