Neverside

quick code check

Can't seem to figure out what is causing an error here.

PHP:

<?php
$user = "****";
$pass = "****";
$db = "***";

$link = mysql_connect("localhost", "$user", "$pass");
if (! $link) 
    die ("couldn't connect to MySQL");
mysql_select_db($db, $link) or die ("couldn't open $db: ".mysql_error());

$result = mysql_query("SELECT * FROM myDB WHERE id=1");
$a_row = mysql_fetch_array($result);
$num_rows =mysql_num_rows($result); 

print "$num_rows Rows \n ";  //prints 1 Rows
print $a_row;               // prints Array

//each of these below cause an error

foreach($a_row as $key => $value) {
    print $key.$value;
    }
    
while ($a_row) { 
 print("yes"); 
}

?>

Last edited by killrobotkill, January 4th, 2004 07:45 PM (Edited 1 times)

foreach($a_row as $key => $value) {

print $key.' - '.$value;

}

ok youve edited it now :P

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Yeah, must have fixed it when you were typing....anyways I was messing around with the while loops and ran into some problems......I am still confused about this.

This crashes my browser:

PHP:

<?php
$a_row = mysql_fetch_array($result);
while ($a_row){
print "yes";
}
?>

This works:

PHP:

<?php
while($a_row = mysql_fetch_array($result){
print "yes";
}
?>

I don't understand why the second crashes my browser.
Is it because the loop is just checking the variable of $a_row, and not the statement of $a_row = mysql_fetch_array($result)

if($a_row){
print "yes";
}

if youre checking the var

while($a_row = mysql_fetch_array($result){

print "yes";

}

if you want to print yes for every result

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