Display text if statement matches - php

I am a beginner when it comes to php, and I have encountered a problem that I cannot find the solution for. I have tried searching for a relevant answer but I haven't found one.
I have the following code in my index.php:
<?php if ($set_status = 2) {
echo 'There is one or more errors';
} else {
echo '';
}
?>
and this a bit further down:
<?php include 'scan/a.php'; ?>
And inside my a.php I have the following code:
<?php
$a = file_get_contents
('http://www.a-random-website/text.html', NULL, NULL, 2, 1);
if ($a == "0") {
include 'fail.php';
} elseif ($a == "1") {
include 'success.php';
} else {
echo 'Offline';
}
?>
And inside my fail.php I have the following code:
<?php
$set_status = 2;
echo 'Failed';
?>
So the idea here is that "a.php" will fetch a number from the website (The correct website has either ["1"] or ["0"] displayed that the code will fetch).
Depending on the result it returns, a.php will include either "fail.php" or "success.php", each containing either a success or a fail-message. If file_get_contents return a 0 I also want fail.php to $set_status = 2; which will cause "There is one or more errors" to be displayed on the front-page (index.php).
The reason that I am using include is that there's going to be a "b.php" and "c.php" and "d.php" and so on, all doing the same thing but fetching data from different pages. I want the success or fail-message to remain easy to edit, without having to edit each and every new x.php file.
So here's where it gets problematic. Everything works beautifully, except for the "There is one or more errors"-message that's supposed to trigger if ($set_status = 2).
I can get as far as the message showing, but when I switch the 1 and 0 in "a.php" (To simulate a specific result) the message will still show. I can't seem to figure it out.
So my question is: What have I done wrong, and what is the correct way to do it?
Thank you in advance!
Best regards,
Marc

use == operator in index.php to compare
<?php if ($set_status == 2) {
echo 'There is one or more errors';
} else {
echo '';
}
?>

<?php if ($set_status ==2) {
echo 'There\'s one or more errors';
} else {
echo '';
}
?>
Just change this code ... see slash\ and == in the echo line .. else of your code id fine

Related

magento code for getIsHomePage - with a twist

Not sure how to explain this but i'm trying to run (2) php if statements together using a getIsHomePage block of code. I need for the div replacement to be true if home page = yes and if "certain page url" = yes. If not either of these two pages then the 'else' statement should be true.
Here is the code block I have now:
<?php if(Mage::getBlockSingleton('page/html_header')->getIsHomePage()) {
echo '<div class="main">';
} else {
echo '<div class="main container">';
}
?>
Any assistance would be appreciated.
Thanks,
Ryan.
You can use the below-mentioned code to check if you are on Home page or viewing your custom URL.
<?php
$myCustomUrl = "...";
if($this->getIsHomePage() || $myCustomUrl ) {
echo 'Either you are on Homepage or your page URL is : ' .$myCustomUrl;
} else {
echo 'Nither you are on Homepage nor your page URL is : ' .$myCustomUrl;
}
?>

Determine Logo link based on URL using PHP

Situation is getting a logo on:
domain.com/special_dir/any_page
or
domain.com/special_dir/any_dir/
to use a link to [domain.com/special_dir/].
Everywhere else on [domain.com/] the logo must a link to [domain.com/]
This is what I have so far.
<?php
$host = $_SERVER['SERVER_NAME'] . $_SERVER['REQUEST_URI'];
if( $host == 'domain.com/special_dir/' ) {
echo '<div"><img src="..."></div>';
} else {
echo '<div"><img src="..."></div>';
}
?>
The logo for [domain.com/special_dir/] only works for [domain.com/special_dir/] URL, no others. I suppose the code it doing what it should, I just don't know how to make it recursive. I did search and read a lot of similar situations but none based on PHP code worked for me.
It is WordPress Multi-site setup and the "special_dir" is a regular sub-directory.
How to correct?
Thanks
Your if ($host == 'domain.com/special_dir/') statement means the special link will be printed for domain.com/special_dir/ only. It excludes everything else, including comain.com/special_dir/any_dir and domain.com/special_dir/any_page.
If think you want ...
if (substr($host,0,22) == 'domain.com/special_dir/') { ... }
This did the trick.
<?php
$url = $_SERVER["REQUEST_URI"];
if (strpos($url, "/special_dir/") === 0) {
echo '<div"><img src="..."></div>';
} else {
echo '<div"><img src="..."></div>';
}
?>

PHP Harmful URL protection

I've made this script, but the 4th line isn't right and I have really no clue how to solve this. I really appriciate if someone helps me. This is my code:
<?php
$url = $_GET["url"];
$badsite = array("http://check.com", "http://hotmail.com");
if($url == $badsite) {
echo "This URL is harmful.";
} else {
echo "Not harmful";
header("Location: " . $_GET["url"]);
}
?>
So the thing which doesn't work is the following line
if($url == $badsite) {
How can I make it so it checks if the GET contains a $badsite?
You don't want to check if the value equals the array, you want to check if it's in the array. Perhaps something like this:
if (in_array($url, $badsite)) {
// ...
}
Side note, you don't need (or want, really) this echo statement:
echo "Not harmful";
header("Location: " . $_GET["url"]);
You might get an error by emitting output before sending a header. But even if you buffer output or in some other way suppress that error, there's no reason to emit output when returning a redirect response. The browser would display it for only an instant, if at all. A redirect by itself is a complete HTTP response, no output is required.
In this case you can use the function in_array:
http://php.net/manual/en/function.in-array.php
<?php
$url = $_GET["url"];
$badsite = array("http://check.com", "http://hotmail.com");
if(in_array($url, $basite)) {
echo "This URL is harmful.";
} else {
echo "Not harmful";
header("Location: " . $_GET["url"]);
}
?>

PHP remember url parameter value using $_SESSION

I made a script that shows the value of "school_id" in url parameter.
http://mywebsite.com/mygrade?school_id=00000
I use $_GET['school_id'] to display the ID number.
<?php echo $_GET['school_id']; ?>
But I what I want is if the parameter "school_id" is empty, I want to display the previous data entered.
Example, the user already browse http://mywebsite.com/mygrade?school_id=00000 then he browse http://mywebsite.com/mygrade?school_id= which id has no value. It will still display 00000 which is the previous ID he used.
I used this code below but doesn't work.. :(
<?php
session_start();
$_SESSION['schoo_id'] = $_GET['school_id'];
if ($_GET['school_id'] === null || $_GET['school_id'] == ""){
echo $_SESSION['schoo_id'];
}
else{
$_GET['school_id'];
}
?>
Anyone who get my point and could help me?
I'm going to break this down line by line, please let me know in the comments if I need to explain anything further:
Self explanatory:
<?php
session_start();
There is a typo here:
$_SESSION['schoo_id'] = $_GET['school_id'];
But! Fixing it won't resolve your problem. What happens if $_GET['school_id'] is not defined/blank? Guess what, $_SESSION['school_id'] is now blank. Obviously you don't want this behavior, so you'll want to only set $_SESSION['school_id'] if $_GET['school_id'] is defined
accessing $_GET['school_id'] will throw an E_NOTICE error if it isn't defined, so you'll want to instead check its existence, rather than checking to see if it is null.
if ($_GET['school_id'] === null || $_GET['school_id'] == ""){
Oh, that typo was intended. Why misspell school though? No need! :)
echo $_SESSION['schoo_id'];
What is this doing? Nothing! No echo, nothing. Just accessing a variable and doing nothing with it.
}
else{
$_GET['school_id'];
}
?>
Here's what your code should look like, or at least I believe is what you intend:
<?php
session_start();
if (isset($_GET['school_id']) && $_GET['school_id'] !== ""){
$_SESSION['school_id'] = $_GET['school_id'];
}
// $_SESSION['school_id'] will be guaranteed to be what $_GET['school_id'] is (if set)
// or whatever it was last time it was defined
// always echo it.
echo $_SESSION['school_id'];
?>
<?php
session_start();
if ($_GET['school_id'] === null || $_GET['school_id'] == ""){
echo $_SESSION['schoo_id'];
}
else{
$_GET['school_id'];
$_SESSION['schoo_id'] = $_GET['school_id']; //here set the session
}
?>
I agree with Salman A, the simplest way:
<?php
session_start();
if (is_int($_GET['school_id'])) $_SESSION['school_id'] = $_GET['school_id'];
// further use $_SESSION['school_id'] for your needs.
?>
what you need to do here is save the GET value in SESSION only if it is set for later use so this should work
<?php
session_start();
if (!isset($_GET['school_id']) || $_GET['school_id'] === null || $_GET['school_id'] == ""){
echo $_SESSION['schoo_id'];
}
else{
$_SESSION['schoo_id'] = $_GET['school_id'];
echo $_GET['school_id'];
}
?>
You almost have it.
<?php
session_start();
if (isset($_GET['school_id']) && trim($_GET['school_id']) !== '') {
// its a fair assumption to make that 'school_id' is intended to be an integer,
// however I will not make that assumption on the OP's behalf.
$_SESSION['school_id'] = $_GET['school_id'];
}
if (isset($_SESSION['school_id']) {
echo $_SESSION['school_id'];
}
else {
echo 'have not entered a school id yet';
}
?>

PHP If and else statement -

I am pulling images and information from a mySQL database and displaying with a few PHP functions. At times all of the information isn't there and I need it to basically display:none; but I can't seem to get it - What am I missing? Here is my display function:
<?php if ($recipe->hassliderimage5 == true) {
$recipe->show_image_carousel5();
} else {
}
?>
And here is the PHP function calling it from the database -
if (trim(mysql_result($this->result,0,"imageCarousel5") != '')) {$this->hassliderimage5 = true;} else {$this->hassliderimage5 = false;}
Here is what I got to work for what I wanted - not sure if it is the best solution or not? I'm still kind of new to PHP.
<?php if ($recipe->hassliderimage5 == true) { ?>
<div id="sliderimageFive" class="item">
<?php
$recipe->show_image_carousel5();
?>
</div>
<?php } ?>
You seem to make the mistake to create your view before collecting the data that has to be displayed. A more logical application structure would be to fetch the data from the database, validate it and then create your view according to the amount and type of data you have.
Also, take a look at this: Why shouldn't I use mysql_* functions in PHP?
I can't comment cause i have not enough reputation, so i try it with an answer.
why you are using the trim on mysql_result($this->result,0,"imageCarousel5") != '' this code will give you either true or false no string you have to trim
Maybe you want to trim the return of mysql_result and then check if its empty
if (trim(mysql_result($this->result,0,"imageCarousel5")) != '') {$this->hassliderimage5 = true;} else {$this->hassliderimage5 = false;}
Try this test:
<?php if (filter_var(recipe->hassliderimage5, FILTER_VALIDATE_BOOLEAN) == true) {
$recipe->show_image_carousel5();
} else {
// do other
}
?>
if return true if the value of 'recipe->hassliderimage5' is "1", "true", "on" and "yes"
Enjoy your code!

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