I have submitted some code to the redirected url and now trying to use this to echo some information out but can't figure out where I am going wrong.
I have the following code:
<?php $login_attempt = mysql_real_escape_string($_GET['login_attempt']);
if ($login_attempt) == '1'{
return 'failed';
}
?>
all I want to do is if the url has $login_attempt=1 I want to return the message 'failed' to the page.
There is no point of escaping anything if it doesn't enter anywhere important (like a database).
<?php
if ($_GET['login_attempt'] == '1') {
echo 'failed';
}
?>
Also, you have a problem in your if statement, that's corrected in the code above. Be sure to include all of the condition inside of parenthesis, and not just one side of the equality check.
Also, if you wish to display something on the screen, you should echo it, not return.
how about:
if ($login_attempt == '1'){
echo 'failed';
}
Try this one. Your error in $login_attempt == '1':
<?php $login_attempt = mysql_real_escape_string($_GET['login_attempt']);
if ($login_attempt == '1'){
echo 'failed';
return false;
}
?>
As others already mentioned you have several problems but the syntax error comes from this:
if ($login_attempt) == '1'{
it should be
if ($login_attempt == '1') {
Dont u think if ($login_attempt) == '1' should be something like this ($login_attempt == '1') Sorry...many others also suggested this :P
At the first, I must tell you that you have a mistake in your IF condition. You typed == outside of ().
In addition, you have to be aware of status of setting your variable through your URL. Check the code below. In this code, I made a function to check the status. Default status is true, and we will check it just for a negative condition. I hope it could be useful for you:
<?php
function check() {
if (isset($_GET['login_attempt'])) {
$login_attempt = mysql_real_escape_string($_GET['login_attempt']);
if ($login_attempt == '1') {
return false;
} else {
return true;
}
} else {
return true;
}
}
if (!check()) echo('Error Message');
?>
Related
Is there any way in PHP to return at else of first statement, if the second statement which is inside of first, is false
if($first == true) {
//other code which is not necessary to mention here!
if($second == true){
// do smth
}
else{
return to the else of $first statement
}
//other code which is not necessary to mention here!
}
else{
//do smth else
}
Yes, there are multiple ways. For starters, just combine both the statements and give another condition:
if ($first == true && $second == true) {
// do smth
} elseif ($first == true && $second == false) {
// else of$first statement
} else {
//do smth else
}
This can be used as a guidance to get an idea to start. But if you can get a real world example, there can be conditions grouped, tailored to your requirement.
While there is no native way to jump to outer elses from an inner else, but you can set a flag for later processing:
$do_else = false;
if($first == true) {
//other code which is not necessary to mention here!
if($second == true){
// do smth
}
else{
$do_else = true;
}
//other code which is not necessary to mention here!
}
else{
$do_else = true;
//do smth else
}
if($do_else){
//do smth else
}
If the answers above doesn t help you in the real situation, you can create a function for execute in 'else' statements to avoid code duplication
I want to check if either of these checkboxes are checked and if either of them are i want the new assigned variable $open_monday to = "yes else "no".
if (isset($_POST['open_monday_lunch'] or $_POST['opening_monday1'])) {
$open_monday = "yes";
}
else { $open_monday = "no"; }
Is that the right way to do it? I have never used or before. I just get a blank pag when trying to run it as if the syntax is incorrect.
Due to Operator precedence its always preferable to use && instead of and ,similarly || instead of or.please refer this link for additional info,
so try like this,
if (isset($_POST['open_monday_lunch'] ) || isset( $_POST['opening_monday1'])) {
$open_monday = "yes";
}
else { $open_monday = "no"; }
You can check it with -
if (!empty($_POST['open_monday_lunch']) || !empty($_POST['opening_monday1'])) {
I'm trying to compare a POST variable with a string. Can someone help me see what in my PHP code is not written correctly? I've tried both '==' and '==='. Thank you for your help.
$action = mysqli_real_escape_string($mysqli, $_POST['action']);
if(strcmp($action, "save") == 0){
//do stuff
}elseif(strcmp($action, "load") == 0){
//do other stuff
}else{
//do even more stuff
}
why not simply use
if ($_POST['action']=='save'){
}elseif($_POST['action']=='load'){
}
don't understand the mysql in this contenxt
Don't know why you want to do this, but try casting $aciton, like (string)$action.
== is used to see if the two sides of comparison are equal, while === is used to check to see if they're identical meaning they are equal AND of the same type.
As for your code, you should just be able to do
if($action == 'save'){
echo 'save';
}
elseif ($action == 'load'){
echo 'load';
}
else{
echo 'none';
}
something is wrong with my code formatting i believe
i am still unsure of what is happening that gives this error,
i am getting the error Parse error: syntax error, unexpected T_VARIABLE, expecting '('
here is my code
<?php
$runamazonapi = false;
if $runamazonapi = true
{
"run this code"
else
}
//do nothing
{
?>
i am getting the following error on line 3 or at this part
if $runamazonapi = true
thanks for your help in advance!!
<?php
$runamazonapi = false;
if ($runamazonapi == true)
{
"run this code"
}
else
//do nothing
{
}
?>
There are a number of syntactical errors with your code, but the error means that the php parser expected to find an ( but instead found a variable. You need () around the if statement condition and you need a closing } on the first if condition. Also, you need to use the proper {} to open and close the else clause:
<?php
$runamazonapi = false;
if ($runamazonapi = true)
{
"run this code"
}
else
{
//do nothing
}
?>
Also, what you have won't work. You're assigning $runamazonapi to true, not checking if it is true. You need to use == not =:
<?php
$runamazonapi = false;
if ($runamazonapi == true)
{
"run this code"
}
else
{
//do nothing
}
?>
try
if($runamazonapi){
//run code
}else{
//do something
}
Somewhere along the line I'm adding or leaving out a { } but I just can't figure out where
<?php
if (file_exists('config.php')) {
require_once('config.php');
{
if ( $EDITED_CONFIG == false )
{
header("Location: welcome.php");
}
}
}
else (file_exists('default-config-new.php')) {
require_once('default-config-new.php');
{
if ( $EDITED_CONFIG == false )
{
header("Location: welcome.php");
}
}
}
?>
If file exists require it and if edited = false redirect, if true end script.
else
If file exists require it and if edited = false redirect, if true end script.
So if the first file doesn't exist it mustn't require it or look for edited, it must skip to the second file and if that exists it must checked edited and then if is false then redirect. If the first file is true it must end script and load page. So it mustn't check second file if first file is true.
Also is this the lightest way to do this?
Thanks
If you indent your code properly, your error will become evident.
A few links that may be useful:
Wikipedia: Indent style
How to indent code
PHP Coding standard: Indentation
You're not closing your if statements: Should be something like:
<?php
if (file_exists('config.php')) {
require_once('config.php');
if ($EDITED_CONFIG == false) {
header("Location: welcome.php");
}
}
else{
require_once('default-config-new.php');
if ($EDITED_CONFIG == false) {
header("Location: welcome.php");
}
}
?>
Edited. Also, you need to close brackets around all code to be executed for that statement, before you can use another elseif or else statement:
if ($x == 1) {
echo "X is 1!";
}
else if ($x == 0) {
echo "X is 0!";
}
else {
echo "Not 1 or 0!";
}
You are missing the { after the else to enclose what you want inside the "else" block i believe
You need to write
else if (conditions...)
You have got
else (conditions...)