Basically, I've been trying to make a simple Wordpress widget that displays a QR code with the URL of the current page. I'm using a modififed version of the simple text widget that parses PHP too.
function the_qrcode($permalink = '', $title = '') {
if($permalink && $title == '') {
$permalink = 'http://eternityofgamers.com/forums';
$title = 'Forums';
}
echo '<img src="http://api.qrserver.com/v1/create-qr-code/?data=' .$permalink. '" alt="QR: ' .$title. '"/>;
}
Can someone tell me what's wrong with this? I get a 500 error when I add it to functions.php.
You will need to use the urlencode() function. Generally as a rule of thumb all querystring values should be url encoded.
function the_qrcode( $permalink = '' ) {
if($permalink == '') {
$permalink = 'http://eternityofgamers.com/forums';
}
echo '<img src="http://api.qrserver.com/v1/create-qr-code/?data='.urlencode($permalink);
}
Now you can create your QR code with:
the_qrcode(the_permalink());
Also, you had a very bad missing equals sign. It is very important to understand the difference between = and ==. If you don't, no matter the context = and == mean two different things. = assigns the right hand side to the left. == returns true or false whether the left and right hand side are loosely equal (loosely because casting will be used if the sides are not of the same type).
Look at this example (Codepad demo):
$a = 5;
$b = 10;
if($a = 6) {
echo "This always appears because when you assign a truthy (all non-zero numbers are true) to a variable, true is returned.\n";
echo "Also a should now equal six instead of five: " . $a . "\n";
}
if($b == 10) {
echo "This will work as expected because == is a comparison not an assignment.\n";
echo "And b should still be 10: " . $b;
}
Try with:
<?php
function the_permalink( $permalink ) {
if ($permalink == '') {
echo '<img src="http://api.qrserver.com/v1/create-qr-code/?data=http://eternityofgamers.com/forums" alt="QR Code">';
} else {
echo '<img src="http://api.qrserver.com/v1/create-qr-code/?data='.$permalink;
}
}
?>
(I've corrected a bunch of syntax errors)
Related
Can Someone tell me what's wrong with this code
function someFunction($num=null) {
return $num;
}
if ($name = someFunction('mystr') && $name ) {
echo 'hello ';
}else {
echo 'hi';
}
Why it is going in else condition and giving me notice of undefined variable $name
Edited: - if i do like this
if($name = someFunction() && $name ){
echo 'hello';
} else {
echo 'hi';
}
this time its also going on else condition as it should but it also not showing the error as i understand it, php just check my first condition $name = someFunction() and its fail then just else
but if i do as i do previously $name = someFunction('str') now $name is set so why notice of undefined variable
sorry for bad grammer
just want to know what is happening here.
It's because logical operators like && have higher precedence than the assignment operator =. You can read more about this here: http://php.net/manual/en/language.operators.precedence.php
The line
if ($name = someFunction('mystr') && $name ) {
is being evaluated like this:
if ($name = (someFunction('mystr') && $name) ) {
where the expression in the inner brackets is evaluated first. Because $name has not been defined before this point, a notice is raised.
I think what you're trying to do is this:
if (($name = someFunction('mystr')) && $name ) {
where you assign the value mystr to $name and then also evaluate that it's "truthy". But as pointed out in the comments, this is a bit of a strange approach. The following code would be equivalent:
$name = 'mystr';
if ($name) {
...
This feels a bit like a problem that's been cut down a bit too much in order to explain it, because it's not really clear why you're doing this.
you code output : hi becuase there is $name is in second is always null.
function someFunction($num=null) {
return $num;
}
$name = someFunction('mystr');
if ($name && $name ) {
echo 'hello ';
}else {
echo 'hi';
}
output : mystr && mystr = 1 then output is hello
You haven't assigned anything to your variable $name and since it is an AND Condition, the first gets true but the second one doesn't so assign any value to your $name and add a condition for it to work.
I have this script:
function DecryptId($id) {
$first_digit = substr($id, 0,1);
if ( $first_digit == 0 ) {
return 'yes';
} else {
return 'no';
}
}
$id = 014;
echo DecryptId($id);
//=> no
Demo
Why it prints no? I expect it prints yes. Because the value of $id starts with 0. What's wrong?
EDIT: In reality I'm passing $id like this: DecryptId($_POST['au']);. $_POST['au'] is containing a number. Something like these:
23
43552
0153
314
09884
As you see, sometimes that number starts with 0. And I need to pass it as a string. How can I do that?
Because of the leading zero, PHP will be parsing that number as octal. Even if it didn't do this, most languages will strip off the leading zeros (since they don't actually form part of the number). This means that $id will evaluate to 12.
Are you sure you don't want to declare it as a string? ($id = "014")
Your function is working fine the issue is that you are passing a number in your function when you should provide a string. So in the case that your variable type is integer the leading zero will eventually fly away.
You can add something to your function to check the variable type and inform the user.
function DecryptId($id) {
$type = gettype( $id );
if($type!= "string") {
echo "Your variable has type ".$type.". Use a 'string' type variable";
return;
}
$first_digit = substr($id, 0,1);
if ( $first_digit == 0 ) {
return 'yes';
} else {
return 'no';
}
}
$id = 014;
echo DecryptId($id);
echo "\n";
$id = '014';
echo DecryptId($id);
Try the above example in PHP Sandbox
try this
<?php
function DecryptId($id) {
$first_digit = substr($id, 0,1);
if ( $first_digit == 0 ) {
return 'yes';
} else {
return 'no';
}
}
$id = '014';
echo DecryptId($id);
?>
For example:
$k = "+";
$q = 8;
echo $array[$q+1];
But I want the following:
echo $array[$q$k1];
So it basically says "call the value of array which is 8+1 so 9." and if I want to call 7 I can do $k = "-".
In PHP, you can not treat operators as variables.
Still, there two basic ways you can achieve the same effect.
You can use a conditional and specify the values accordingly:
$r = 1; //the value you're adding; moved to a variable for clarity
$op = '+'; //+ means add; anything else means subtract
echo $array[$q + ($op === '+' ? $r : -$r)];
//or
if($op === '+') {
echo $array[$q + $r];
} else {
echo $array[$q - $r];
}
Or you can change the operation into a multiplication:
echo $array[$q + (($op === '+' ? 1 : -1) * $r)];
Either form will work; it's just a matter of what is most convenient for your code.
I'm outputting a custom user field in Wordpress, which is user's Twitter username. Some users might add it with '#' symbol, some might without. How can I output that field and check if the symbol already exists, if not, add it?
This is code I'm using to output usrename:
<?php echo get_the_author_meta('twitter_name'); ?>
<?php
$authorMeta = get_the_author_meta('twitter_name');
if (strpos('#', $authorMeta) !== 0) {
$authorMeta = '#'.$authorMeta;
}
echo $authorMeta;
you need to check if # is on 1st place, you can do this in many ways
<?php
$authorMeta = get_the_author_meta('twitter_name');
if ($authoMeta[0] != '#') {
$authorMeta = '#'.$authorMeta;
}
echo $authorMeta;
Try this...it might help
$a = get_the_author_meta('twitter_name');
$b = explode('#',$a);
if($b[1] == 0){
echo "#".$a;
}
else{
echo $a;
}
Having one of those brain fade moments this morning. I have the following php:
$imgset = $result->fields[6];
if ($imgset = '')
{
$imgset = 'logo';
}
else
{
$imgset = $result->fields[6];
}
echo '<img id="imgdisp" src="/img/'.$imgset.'.jpg" />';
$imgset = '';
Where it looks to see if the result from the database is blank and if so, should put in logo.jpg instead of whatever the result is. For some reason though, it just does not want to work and I am probably being completely blind, but cannot see why not. I still get blank images in the HTML and filenames of "/img/.jpg" as though $imgset is still passing through a blank. The values are not NULL in the SQL either, they are most definitely blank entries inputted from an inputbox using a _POST in a form elsewhere.
This:
if ($imgset = '') {
Is always setting $imgset to empty. Use comparison instead:
if ($imgset == '') {
Your else is also not needed since in that case $imgset is already set as $result->fields[6];.
Try to verify if the image exists in your path as well
<?php
$imgset = $result->fields[6];
if ($imgset) {
$imgset = $result->fields[6];
$path ='pathtoimages';
if(!file_exists($path.'/'.$imageset.'.jpg'){
$imgset = 'logo';
}
}
else
{
$imgset = 'logo';
}
echo '<img id="imgdisp" src="/img/'.$imgset.'.jpg" />';
?>
You forgot to compare on the if condition and instead you are assigning an empty value to $imgset. if ($imgset = '') should be if ($imgset == '')
$imgset = $result->fields[6];
if ($imgset == '')
{
$imgset = 'logo';
}
echo '<img id="imgdisp" src="/img/'.$imgset.'.jpg" />';
$imgset = '';
you do not need the else part as the value is already assigned in the first statement.
Using a ternary operator it can be done like this:
echo '<img id="imgdisp" src="/img/'.(empty($imgset)?'logo':$imgset).'.jpg" />';
Shorter code at the cost of readability.
This is the reason why it's better to reverse the condition:
if ('' = $imgset)
would have lead to an error.
The answer:
if ('' == $imgset)
//or
if (empty($imgset))
If you are selecting from MYSQL, you can use something like
SELECT *,COALESCE(image,"logo") AS image FROM ....
This way when the results come back and some rows have a NULL image, it will be replaced by "logo" so you don't need the IF logic in your PHP :)