]Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result

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]Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result

priyankagound

Hello Experts,

             I am stuck in database query. Well i am making a page which retrieve my database value.

For which below is my code:-

<?php 
$con=mysqli_connect("localhost","root","","test"); 
if (mysqli_connect_errno()) 
{ 
echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
}

$result = mysqli_query($con,"SELECT * FROM testing where flag=1 ");

foreach ($result as $results){

while($row = mysqli_fetch_array($results)) 
{ 
echo  $row['review'] ;
echo  $row['name'] ;
} 
}
mysqli_close($con); 
?>

When i run the above code it shows me error:

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, array given in /home/stradsol/public_html/test1.php on line 14

I also tried by replacing with:

while($row = mysqli_fetch_array($results, MYSQL_BOTH))

But still the same error

I am stuck. I dont know what I am missing.

Any help is appreciated.

ginerjm

Having a bad day?

YOu are using a foreach to break out a single row from the query results resource/array. But then you feed a single row to the while loop which doesn't make much sense.

Dump the foreach.

priyankagound

ginerjm;1305621 wrote:
Having a bad day?

YOu are using a foreach to break out a single row from the query results resource/array. But then you feed a single row to the while loop which doesn't make much sense.

Dump the foreach.

Even after removing the foreach loop it showing the same "warning" .
Actually i have to use some css in each of my review row records therefore i am using "foreach loop" .
So please can some one help me out

ginerjm

You CAN'T use a foreach and a while loop on the results of a query!!! The message is telling you exactly what is wrong - why can't you read it?

And what the H... does a foreach loop have to do with CSS that your while loop won't accomplish?

Drop the foreach and use this line:

while ($row = mysqli_fetch_array($result))

Trust me!

priyankagound

ginerjm;1305743 wrote:
You CAN'T use a foreach and a while loop on the results of a query!!! The message is telling you exactly what is wrong - why can't you read it?

And what the H... does a foreach loop have to do with CSS that your while loop won't accomplish?

Drop the foreach and use this line:

while ($row = mysqli_fetch_array($result))

Trust me!

I trust u and the code runs exactly perfect. But i want to apply css to each of my records as in "quote css". For that what can i do ??

I'l paste my css done code:


<?php 
$con=mysqli_connect("localhost","root","","test"); 
if (mysqli_connect_errno()) 
{ 
echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
}


$result = mysqli_query($con,"SELECT * FROM testing where flag=1 ");

foreach ($result as $results){
echo "<div class=\"testimonial\">";
while($row = mysqli_fetch_array($results, MYSQL_BOTH)) 
{ 

echo  $row['review'] ;
echo "<div class=\"arrow-down\">";
echo "</div>";
echo "<div class=\"testimonial-author\">";
echo  $row['name'] ;
echo "</div>";
} 

echo "</div>"."<br>";

}


mysqli_close($con); 
?>

Is der any other way?? what must i do now
I am stuck and really my brains not working for this.

ginerjm

ok - I give up.

I've told you twice YOU CAN'T USE FOREACH AND WHILE, but you insist that you know better.

Good luck.

priyankagound

ginerjm;1305749 wrote:
ok - I give up.

I've told you twice YOU CAN'T USE FOREACH AND WHILE, but you insist that you know better.

Good luck.

I dont know it therefore i am here with my doubts. I really appreciated while people are correcting me with my mistake and so do i appreciate your reply.
Please correct me with my code. Paste the code so that can help me out.

ginerjm

Do What I said to do.

As for adding styling - So far you are just echoing out two data items. Change those lines to some real html code showing that data and include whatever css you want to in those lines. So simple. Try it. Then show us YOUR code.

NogDog

Note that mysqli_query() never returns an array (nor an iterable object), so using its result in a foreach() will never work. Here's a somewhat more robust version that should at least tell you where any errors are (though if I were doing it I'd use the object-oriented style, not the functional style):

<?php 
$con=mysqli_connect("localhost","root","","test"); 
if (mysqli_connect_errno()) 
{ 
	die("Failed to connect to MySQL: " . mysqli_connect_error()); 
}
// define query separately so we can reference it in error message:
$query = "SELECT * FROM testing where flag=1";
$result = mysqli_query($con, $query);
// always make sure you get a valid result!!!
if($result == false) {
	die("Query failed: ".mysqli_error().PHP_EOL.$query);
}
// might as well limit results to associative array since that's how you're using it:
while($row = mysqli_fetch_assoc($result)) // $result, not $results 
{ 
        // example of using HTML in output:
	echo  "<h3>".$row['review']."</h3>\n";
	echo  "<p>".$row['name']."</p>\n";
} 
mysqli_close($con);

priyankagound

NogDog;1305759 wrote:
Note that mysqli_query() never returns an array (nor an iterable object), so using its result in a foreach() will never work. Here's a somewhat more robust version that should at least tell you where any errors are (though if I were doing it I'd use the object-oriented style, not the functional style):
<?php 
$con=mysqli_connect("localhost","root","","test"); 
if (mysqli_connect_errno()) 
{ 
	die("Failed to connect to MySQL: " . mysqli_connect_error()); 
}
// define query separately so we can reference it in error message:
$query = "SELECT * FROM testing where flag=1";
$result = mysqli_query($con, $query);
// always make sure you get a valid result!!!
if($result == false) {
	die("Query failed: ".mysqli_error().PHP_EOL.$query);
}
// might as well limit results to associative array since that's how you're using it:
while($row = mysqli_fetch_assoc($result)) // $result, not $results 
{ 
        // example of using HTML in output:
	echo  "<h3>".$row['review']."</h3>\n";
	echo  "<p>".$row['name']."</p>\n";
} 
mysqli_close($con); 

Thanks for your reply. It has really helped me.
I tried out with the below code and my problem is now solved.

<?php 
$con=mysqli_connect("localhost","root","","test"); 
if (mysqli_connect_errno()) 
{ 
echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
}
$result = mysqli_query($con,"SELECT * FROM testing where flag=1 ");
while($row = mysqli_fetch_array($result, MYSQL_BOTH)) 
{ 
echo "<div class=\"testimonial\">";
echo  $row['review'] ;
echo "<div class=\"arrow-down\">";
echo "</div>";
echo "<div class=\"testimonial-author\">";
echo  $row['name'] ;
echo "</div>";
echo "</div>"."<br>";
} 
mysqli_close($con); 
?>

Thanks a lot all.

williamforrest

You cannot do a foreach on the return value of mysqli_query.
And there's no mysqli_query only mysql_query.

NogDog

williamforrest;1305839 wrote:
You cannot do a foreach on the return value of mysqli_query.
And there's no mysqli_query only mysql_query.

Hmm...maybe you should write a bug report against the php manual page for mysqli_query(), then?

pisit_sae-ngow

used followme can resolve this problem.See it for example.

<?php
include("connect.php");

$sql = "SELECT id, event, sensor1, sensor2, sensor3 FROM tablearduino";
$result = $conn->query($sql);
if ($result->num_rows > 0) 
{   // output data of each row
   while($row = $result->fetch_assoc()) 
   {
	echo"<tr>"; 
		echo "<td>".$row["id"]."</td>";
		echo "<td>".$row["event"]."</td>";
		echo "<td>".$row["sensor1"]."</td>";
		echo "<td>".$row["sensor2"]."</td>";
		echo "<td>".$row["sensor3"]."</td>";
	echo"</tr>";
   }
} else 
	{
	echo "0 results";
	}

NogDog

sigh

Why do people insist on digging up old, old threads like that?

Closing....

sigh