为什么外部变量超出了PHP的作用范围？ [重复]

Question

This question already has an answer here:

• Why are my variables going out of scope? PHP 2 answers

I have a small problem with a PHP function that can't access a variable that was defined outside its code block and I can't see what the problem is. Redefining the variable inside the function block works fine but seems a superfluous step:

<?php
    //constants for database connection
    define(DB_RESOURCE, "192.168.99.67");
    define(DB_USERNAME, "myUsername");
    define(DB_PASSWORD, "p@ssword");
    define(DB_NAME, "myDatabase");
    //variable defined here so why can't function see it?
    $dbConnect = mysqli_connect(DB_RESOURCE, DB_USERNAME, DB_PASSWORD, DB_NAME);


    function readData($table, $column) {
        $sql = "SELECT {$column} FROM {table}";
        //redefining this solves the problem but why is this needed?
        // $dbConnect = mysqli_connect(DB_URL, DB_USER, DB_PASS, DB_NAME);
        //the $dbConnect variable is not defined and error is flagged in PHPStorm in the line below
        $data =  mysqli_query($dbConnect, $sql);
    }
?>

Any ideas why this is happening?

Many thanks for your help,

Kw

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Answer 1

in php if you want to use any variable inside any function without passing it in parameter. you need to defined it as global. it before using you need global $dbConnect.

<?php
    //constants for database connection
    define(DB_RESOURCE, "192.168.99.67");
    define(DB_USERNAME, "myUsername");
    define(DB_PASSWORD, "p@ssword");
    define(DB_NAME, "myDatabase");
    //variable defined here so why can't function see it?
    $dbConnect = mysqli_connect(DB_RESOURCE, DB_USERNAME, DB_PASSWORD, DB_NAME);


    function readData($table, $column) {
        $sql = "SELECT {$column} FROM {table}";
        //redefining this solves the problem but why is this needed?
        global  $dbConnect;
        // now you can use $dbConnect. In php you need to specify it as global once inside the block after it you can use 

        $data =  mysqli_query($dbConnect, $sql);
    }
?>