I am trying to set a flag to show or hide a page element, but it always displays even when the expression is false.
$canMerge = ($condition1 && $condition2) ? 'true' : 'false';
...
<?php if ($canMerge) { ?>Stuff<?php } ?>
What's up?
This is broken because 'false' as a string will evaluate to true as a boolean.
However, this is an unneeded ternary expression, because the resulting values are simple true and false. This would be equivalent:
$canMerge = ($condition1 && $condition2);
The value of 'false' is true. You need to remove the quotes:
$canMerge = ($condition1 && $condition2) ? true : false;
Seems to me a reasonable question especially because of the discrepancy in the way PHP works.
For instance, the following code will output 'its false'
$a = '0';
if($a)
{
echo 'its true';
}
else
{
echo 'its false';
}
You are using 'true' and 'false' as string. Using a string(non-empty and not '0' and not ' ', because these are empty strings and will be assume as false) as a condition will results the condition to be true.
I will write some correct conditions that could be use:
$canMerge = ($condition1 && $condition2);
$canMerge = ($condition1 && $condition2) ? true : false;
$canMerge = ($condition1 && $condition2) ? 'true' : 'false';
...
<?php if ($canMerge == 'true') { ?>Stuff<?php } ?>
$canMerge = ($condition1 && $condition2);
then
if ($canMerge){
echo "Stuff";
}
Related
I have array look like this.
$setLogic = array(array("Conj"=>null,"Topic"=>True),array("Conj"=>"Or","Topic"=>True),array("Conj"=>"Or","Topic"=>false),
I try to build a IF statement with the dynamicly with this way.
foreach($setLogic as $value){
echo $value['Conj'].(int)$value['Topic'];
if($value['Conj'].(int)$value['Topic'] == true){
$getBoolean[] = true;
}
else{
$getBoolean[] = false;
}
I just need something like this.
(true or true or false) and return true or false
I found my answer with my self.
It's look like this.
$abc= '';
foreach($setLogic2 as $key => $value){
$abc.= $value['Conj'].' '.(int)$value['Topic'].' ';
}
//$abc = "0 And 1 And 1";
if(eval("return $abc;") == true) {
echo 'true boolean';
}
else if(eval("return $abc;") == false){
echo 'false boolean';
}
else{
echo 'none';
}
Ok i will re formulate my answer....
You want to dynamically evaluate an array which contains for each subarray
* a verb : "and" or "or"
* a value "true or "false"
Quicker and most dirty way is to use eval().
* build your conditions as a string (replace "and" by "&&" and "or" by "||"
'true || true && false'
then put the result in a variable and evaluate
eval('$result = (true || true && false);');
var_dump($result);
This works for sure...
I have a var which return TRUE or FALSE. how do I use ternary logic to return the string
"YES" if $var1 is TRUE and
"NO" if $var1 is FALSE ?
$Var1 = TRUE; /*dynamic value*/
$status = ($Var1 == true ? $Var1:"NO");
Thank you.
$status = $Var1 ? 'YES' : 'NO';
You cannot go shorter than that :)
you do not need neither () nor == TRUE since if $var1 returns true if it is not null, 0 or false.
$status = ($Var1 == true ? "YES":"NO");
$status = ( $var1 === true ) ? "YES" : "NO";
I need to display true or false not the number 1
$permission === 1 ? true: false
nearly right:
$permission= ($permission === 1) ? 'true': 'false';
echo $permission;
http://php.net/manual/en/language.operators.comparison.php
First you dont make any assignment =
then you have to make a comparison $permission === 1
then you want the string values "true" or "false"
change the boolean to string value
$permission === 1 ? 'true': 'false';
Its as simple as
echo "Output : ".($permission == 1) ? 'true': 'false';
I am running into a funny problem with a mischievous "if" condition :
$condition1="53==56";
$condition2="53==57";
$condition3="53==58";
$condition=$condition1."||".$condition2."||".$condition3;
if($condition)
{
echo "blah";
}
else
{
echo "foo";
}
Why does the if condition pass?
Why does php echo "blah"? What do I do to make php evaluate the "if" statement and print "foo"?
The problem here is that you're putting your expressions in strings!
Your $condition1, $condition2, and $condition3 variables contain strings, and not the result of an expression, and the same goes for your $condition variable which will be a string that looks like 53==56||53==57||53==58. When PHP evaluates a string it considers it true if it is not empty and not equal to 0, so your script will output blah.
To fix this you just need to take your expressions out of the strings. It should look like this:
$condition1 = 53 == 56; // false
$condition2 = 53 == 57; // false
$condition3 = 53 == 58; // false
$condition = $condition1 || $condition2 || $condition3; // false || false || false = false
if ($condition) {
echo 'blah';
} else {
echo 'foo'; // This will be output
}
You're evaluating strings as booleans; they'll aways be true (except the strings "" and "0". Get rid of almost all of the quotes in your program.
Those aren't conditions, they're strings.
$condition1=53==56;
$condition2=53==57;
$condition3=53==58;
$condition=$condition1 || $condition2 || $condition3;
if($condition)
{
echo "blah";
}
else
{
echo "foo";
}
Because you're not checking those variables, it's saying if (String) will always return true. (unless "")
You should be doing:
if(53==56 || 53==57 || 53==58)
{
echo "blah";
}
else
{
echo "foo";
}
All $condition* variables will evaluate to true. This is how PHP sees it:
if("53==56" || "53==57" || "53==58")
What you want is this:
$condition1 = 53==56;
$condition2 = 53==57;
$condition3 = 53==58;
It's because you're evaluating a string, and strings other than empty strings evaluate to true.
You are concatting a string together, a non-empty string equals TRUE in php.
Because when the if passes, $condition is a string (a concatenation of) containing the text of your conditions. Try using if(eval($condition)).
String always evaluate to true if its not empty
And btw php make implicit conversion to boolean
Is there a function to check both
if (isset($var) && $var) ?
The empty() function will do the job.
Use it with the not operator (!) to test "if not empty", i.e.
if(!empty($var)){
}
You may use the ?? operator as such:
if($var ?? false){
...
}
What this does is checks if $var is set and keep it's value. If not, the expression evaluates as the second parameter, in this case false but could be use in other ways like:
// $a is not set
$b = 16;
echo $a ?? 2; // outputs 2
echo $a ?? $b ?? 7; // outputs 16
More info here:
https://lornajane.net/posts/2015/new-in-php-7-null-coalesce-operator
there you go. that should do it.
if (isset($var) && $var)
if (! empty($var))
It seems as though #phihag and #steveo225 are correct.
Determine whether a variable is considered to be empty. A variable is
considered empty if it does not exist or if its value equals FALSE.
empty() does not generate a warning if the variable does not exist.
No warning is generated if the variable does not exist. That means
empty() is essentially the concise equivalent to !isset($var) || $var
== false.
So, it seems !empty($var) would be the equivalent to isset() && $var == true.
http://us2.php.net/empty
Try the empty function:
http://us2.php.net/empty
isset($a{0})
isset AND len is not 0 seems more reliable to me, if you run the following:
<?php
$a=$_REQUEST['a'];
if (isset($a{0})) { // Returns "It's 0!!" when test.php?a=0
//if (!empty($a)) { // Returns "It's empty!!" when test.php?a=0
echo 'It\'s '.$a;
} else { echo 'It\'s empty'; }
?>
$a = new stdClass;
$a->var_false = false;
$a->var_true = true;
if ($a->notSetVar ?? false) {
echo 'not_set';
}
if ($a->var_true ?? false) {
echo 'var_true';
}
if ($a->var_false ?? false) {
echo 'var_false';
}
This way:
if (($var ?? false) == true) {
}
I am amazed at all these answers. The correct answer is simply 'no, there is no single function for this'.
empty() tests for unset or false. So when you use !empty(), you test for NOT UNSET (set) and NOT FALSE. However, 'not false' is not the same as true. For example, the string 'carrots' is not false:
$var = 'carrots'; if (!empty($var)){print 1;} //prints 1
in fact your current solution also has this type problem
$var = 'carrots'; if (isset($var) && $var){print 1;} //prints 1
as does even this
$var = '1.03'; if (isset($var) && $var == true){print 1;} //prints 1
in fact... if you want to do as you described exactly, you need:
$var = 'carrots'; if (isset($var) && $var === true){print 1;} //Note the 3 Equals //doesn't print 1
I suppose the shortest valid way to test this case is :
if (#$var === true){ print 1;}
But suppressing errors for something like this is pretty awful practice.
Don't know if an exact one already exists, but you could easily write a custom function to handle this.
function isset_and_true($var) {
return (isset($var) && $var == true) ? true : false;
}
if (isset_and_true($a)) {
print "It's set!";
}
Check if the variable is set, and true. Ignore warning message
if(#!empty($foo))