I have this JS inside my php code:
echo " for (var i = 0; i<length; i++){
alert('array[i]');
}";
Assuming all variables were defined and initialized, i'm not getting any output from the alert.
However, if i replace array[i] with array[2], I get that value alerted.
any advice?
Updated variant:
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$storeArray[] = $row['DayNum'];
}
$length = count($storeArray);
for($i=0; $i < $length; $i++) {
echo "alert(".$storeArray[$i].");";
}
You need a script tag..Its not possible to alert something the way you are doing it.
<script>//write your javascript here</script>
Example:
<?php
function alert($myArray)
{
echo '<script type="text/javascript">alert("' . $myArray . '"); </script>';
}
?>
Can it be length is not defined and you think it's giving you the array lenght? in that case you should have i < array.length otherwise length is thought to be a variable.
Check also that array[i] being an "array" where you get the values from in the for loop you don't need the '. Just write alert(array[i]);
alert('array[i]') in this part JavaScript not execute array[i] as variable, instead, it prints it as string because its enclosed with single quota, change it to:
echo '<script>';
echo '
for (var i = 0; i<length; i++){
alert(array[i]);
}
';
echo '</script>';
<?php echo "<script> for(var i=0; i<array.length; i++){alert(array[i]);} </script>"; ?>
