mysql_fetch_array() expects parameter 1 to be resource, string given

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while($result = mysql_fetch_array($query)){
            echo "<p>Departure:".$result[0]."</p>";
            echo "<p>Arive:".$result[1]."</p>";
            echo "<p>Date:" .$result[2]."</p>";
            echo "<p>Time:".$result[3]."</p>";

        }

Posted 31-Aug-14 2:28am

JaironLanda

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Nelek 31-Aug-14 8:50am

?????????

Don't think we can read minds or do astral projections to see your monitor. If you need help, the least you could do is to add some relevant code to your question or to explain your problem in such a way, that the users of CP can understand it. Otherwise, nobody will be able to help you.

Please use the "improve question" and add relevant information or a piece of code envolved. There are thousands of ways to screw things up, how are we supposed to know which one did you choose?

Have a look to:
What have you tried?[^]
How to ask a question?[^]

Mohibur Rashid 31-Aug-14 9:46am

did you follow proper steps?
example:
$query="SELECT * FROM sometable";
$result=mysql_query($query, $conn);
$row=mysql_fetch_row($result);

2 solutions

Solution 1

It means that query failed or returning an error.

check the query properly, echo that query variable and run it in mysql and check the mysql error.

Posted 31-Aug-14 2:46am

pandu web dev

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ChauhanAjay · Accepted Answer · 2014-08-31T04:46:00

Try this code maybe it helps

PHP

<?php
$con = mysql_connect("yourhostname","youruserid","password");
mysql_select_db("world");
$result = mysql_query("SELECT * FROM Country",$con) or die(mysql_error());
echo mysql_num_rows($result);
while($row = mysql_fetch_array($result)) {
  echo $row['Code'] . "---" . $row['Name'];
  echo "<br>";
} 
?>
</br>