A variable is considered empty if it does not exist or if its value equals FALSE. empty() empty ( mixed$var ) : bool Code language: PHP ( php )ĭetermine whether a variable is considered to be empty. We’ll go over why that’s important later in the article.īefore I discuss the difference and show a few examples, here are the descriptions for empty(), isset(), and is_null() from the php.net manual. empty() and isset() are language constructs, while is_null() is a standard function. Built-in Variable Testing ToolsĪll three of these functions are built into PHP, so they should always be available for your use when writing code. Three of these functions that are easy to mix up are isset(), empty(), and is_null(). There are a variety of functions made to test the state and value of variables, including ones that can tell you if there is anything available to use at all. When I looked into it, it turns out that I was using the wrong function to test for a variable in PHP. This is one of those PD problems, where I wrote some code that stopped functioning. If you just use null != $variable without prechecking for existence, you will mess up your logs with warnings because that is an attempt to read the value of an undefined variable.I’ll be honest: most of the posts that I write are either because I’ve solved a problem for a client, or because I solved a problem that Past-david created. That is aquivalent to array_key_exists('variable', get_defined_vars()) & null != $variable. In most cases, isset($variable) is appropriate. try same non-null variable value as globally defined as well ( array_key_exists('test', get_defined_vars()) test in local scope (what is working in global scope as well)Įcho '$test '. set a global variable to test independence in local scope To achieve an equivalent behaviour in PHP, you can check whether the variable name exists in the keys of the result of get_defined_vars(). IsTouch = createTouch != undefined // true JavaScript's 'strict not equal' operator ( !=) on comparison with undefined does not result in false on null values. This is my personal belief as to why the core PHP developers left isset() to return false when something is null. However, if you're finding that in your code you have to check for whether a variable has been defined or not, then you're likely doing something wrong. $isset = array_key_exists('otherVariable', get_defined_vars()) $isset = array_key_exists('variable', get_defined_vars()) We still can't use isset(get_defined_vars()) here because the key could exist and the value still be null, so we have to use array_key_exists('variable', get_defined_vars()). Using get_defined_vars() will return an associative array with keys as variable names and values as the variable values. So how do you actually check if a variable is defined? You check the defined variables. $variable is being defined as null, but the isset() call still fails. $isset = isset($variable) īut in this example, it won't work like JavaScript's undefined check. In the following example it will work the same way as JavaScript's undefined check. There is no equivalent to JavaScript's undefined (which is what was shown in the question, no jQuery being used there). To check if a variable is undefined you will have to check if the variable is in the list of defined variables, using get_defined_vars(). I don't know why it was decided that isset() would return false if the value is null. It's important to realize in programming that null is something. I don't know why there are so many answers stating that isset() is the way to go, or why the accepted answer states that as well. It seems you've specifically stated that you're not looking for isset() in the question. The isset() function does not check if a variable is defined.
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