CODEWITHSUNDEEP
php
Nagri
Nagri

Reputation: 3136

mysql_fetch_array() not spiting anything

Here is my code:

$campagin_id = $_SESSION['campagin_id_for_camp'];

$query = "SELECT * FROM survey_result where campagin_id = ".$campagin_id;

$conn=mysql_connect($dbconfig['db_hostname'],$dbconfig['db_username'],$dbconfig['db_password']) or die(mysql_error());

mysql_select_db($dbconfig['db_name'],$conn);

$exec_query =mysql_query($query) or die(mysql_error());

$row=mysql_fetch_array($exec_query);
echo "<br> row = ".$row;
while ($row=mysql_fetch_array($exec_query)){
    echo "I am In";

}

The Problem is that I am not getting anything in $row I cant get into the while loop, nothing shows up when I try to echo the value of $row, No error Nothing. Can you help me to find a problem in my code ?

Ps : The database is their. I have checked for the query for the corresponding value of $campagin_id. and also when i tried to echo $exec_query it echoed this : Resource id #8

PPS : The database have more than 7 record for each id so it doesn't matter if I call mysql_fetch_array($exec_query) more than once before going in to the while loop. and for the $campagin_id in the session their are many records present in the database.

Upvotes: 0

Views: 104

Answers (3)

Pankit Kapadia
Pankit Kapadia

Reputation: 1579

You have written $row=mysql_fetch_array($exec_query) and then you are echoing something. and you are using the same in while.

Instead of: 

$row=mysql_fetch_array($exec_query);
echo "<br> row = ".$row;
while ($row=mysql_fetch_array($exec_query)){
    echo "I am In";
}

Use this (as per my knowledge you should not use $row=mysql_fetch_array() once you have used before while):

while ($row=mysql_fetch_array($exec_query)){
    echo "I am In";
}

Upvotes: 3

Константин Рекунов
Константин Рекунов

Reputation: 457

Try this code.

<?
$campagin_id = $_SESSION['campagin_id_for_camp'];
$query = "SELECT * FROM survey_result where campagin_id = ".$campagin_id;

mysql_connect($dbconfig['db_hostname'],$dbconfig['db_username'],$dbconfig['db_password']) or die(mysql_error());
mysql_select_db($dbconfig['db_name']);

$exec_query =mysql_query($query) or die(mysql_error());
while ($row=mysql_fetch_assoc($exec_query)) {
    echo "<br/> row = <pre>".print_r($row)."</pre><br/>";
}
?>

Upvotes: -1

MrCode
MrCode

Reputation: 64526

If the query returns Resource id #8 then that means it was successful - ie there were no errors. There were probably no rows returned by that query, so no rows in your table that match the given campagin_id.

You are also calling mysql_fetch_array() twice separately, you shouldn't do that because your while loop will skip the first row because calling this moves the pointer in the result set forward by one.

Also you can't echo an array as you are trying to, if you want to see the contents of an array use print_r() or var_dump().

I suggest adding some code to handle no rows found:

if($exec_query && mysql_num_rows($exec_query) > 0)
{
    while ($row=mysql_fetch_array($exec_query)){
        echo "Row: " . print_r($row, true);
    }
}
else
{
    echo 'None found';
}

Upvotes: 2

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