| SitePoint Sponsor |
PHP Code:<?php
$db_host = "1234";
$db_user = "1234";
$db_pwd = "1234";
$db_name = "1234";
mysql_connect($db_host, $db_user, $db_pwd);
mysql_select_db($db_name);
?>
<?php
if ( $_POST['lock'] == 'LOCKED' ) {
mysql_query("update authorize set lockedstatus='LOCKED'");
} elseif ( $_POST['lock'] == 'UNLOCKED'] {
mysql_query("update authorize set lockedstatus='UNLOCKED'");
echo "You have sucessfully switched it to $_POST['lock'] status";
?>
no ) on the elseif, instead you used a ]
I also don't see a closing '}'.
Its still not working.
Yep, so to sum up:
PHP Code:if ( $_POST['lock'] == 'LOCKED' ) {
mysql_query("update authorize set lockedstatus='LOCKED'");
} elseif ( $_POST['lock'] == 'UNLOCKED'] ) {
mysql_query("update authorize set lockedstatus='UNLOCKED'");
}
echo "You have sucessfully switched it to $_POST['lock'] status";
It's still blowing up somewhere guys.
Do you have an error message? You're not checking to see if $_POST['lock'] is set before you do a compare. That can cause an error, too.
PHP Code:if (isset($_POST['lock'])) {
$var = $_POST['lock'];
if ( $var == 'LOCKED') {
mysql_query("update authorize set lockedstatus='LOCKED'");
} elseif ( $var == 'UNLOCKED') {
mysql_query("update authorize set lockedstatus='UNLOCKED'");
}
echo "You have sucessfully switched it to $_POST['lock'] status";
}
Try adding this too:Originally Posted by Andrewaclt
PHP Code:if (isset($_POST['lock'])) {
$var = $_POST['lock'];
if ( $var == 'LOCKED') {
mysql_query("update authorize set lockedstatus='LOCKED'") or die("Query Failed: ".mysql_error());
} elseif ( $var == 'UNLOCKED') {
mysql_query("update authorize set lockedstatus='UNLOCKED'") or die("Query Failed: ".mysql_error());
}
echo "You have sucessfully switched it to $_POST['lock'] status";
}
well I echoed the post variable and it showed the correct data. What do you mean if the variable is set?
see above, fixed code.
Andrew, that bombed out.
jeanco that bombed too
No error messages displayed
PHP Code:if (isset($_POST['lock'])) {
$var = $_POST['lock'];
echo $var;
if ( $var == 'LOCKED') {
mysql_query("update authorize set lockedstatus='LOCKED'") or die("Query Failed: ".mysql_error());
} elseif ( $var == 'UNLOCKED') {
mysql_query("update authorize set lockedstatus='UNLOCKED'") or die("Query Failed: ".mysql_error());
}
echo "You have sucessfully switched it to $_POST['lock'] status";
} else {
echo "Lock not set";
}
blank page.
If you get a blank page then something really bad is wronglook in your file system and see if it's logging errors to a file somewhere. like in htdocs or whatever your folder is when you first log in.
I don't have a file system I don't think. Do I need something else in the if thens statement to contact it with the database information?
Here is the page leading up to the if then statement:
PHP Code:<form action="locked.php" method="post">
Would you like to lock or unlock?<select size="1" name="lock">
<option value="UNLOCKED">Unlock</option>
<option value="LOCKED">Lock</option>
</select><br>
<input type="submit" name="submit" value="submit" />
</form>
put error_reporting(E_ALL); in the beginning of your file and see what error you have on the browser or the error log.
No errors were displayed.
Look at to where you upload the php files, look for an error log, on my shared host when I first login I login to a folder containing public_html, etc in this folder php errors are logged. See if there is something similar.
I don't have anything like that unfortunately. Can someone try to replicate this and see if it works for you?
Did you have error_reporting turned on? eg: display_errors On
you can check by making simple page test.php with this code:
<?php phpinfo(); ?>
It's not turned on.
Posting Permissions
Bookmarks