CODEWITHSUNDEEP
phpjsonparsing
CodeCrack
CodeCrack

Reputation: 5373

if() doesn't print out json data unless I use var_dump in php

$json = file_get_contents($url);
$json_output = json_decode($json, TRUE);
var_dump($json_output);
if($json_output){
    echo "TRUE";
} else {
    echo "FALSE";
}

the data from var_dump($json_output) is

boolean true

followed by TRUE from my if statement unless I remove var_dump($json_output), then nothing happens and TRUE does not get printed out either. Why does this happen?

Upvotes: 1

Views: 263

Answers (1)

webbiedave
webbiedave

Reputation: 48887

One could ascertain that when you removed the var_dump line you committed an editing mistake, which caused a syntax error (or the like) and since you have error reporting turned off, you don't see any output.

var_dump does not affect the value of a variable and, therefore, does not affect its conduct in a conditional.

$json = json_encode(true);
$json_output = json_decode($json, TRUE);
//var_dump($json_output);
if($json_output){
    echo "TRUE";
} else {
    echo "FALSE";
}

// output: TRUE

Upvotes: 1

Related Questions