douchun1859 2019-04-28 05:11
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如何使用AJAX调用PHP函数,并获取返回值?

I want to call a specific function from my database.php file, and get the returned value. Here I am trying to do this:

js:

function submit_verification_code(){
$.ajax({
url: "database.php",
type: "post",
data: ({
    'code': code_entered,
}),
dataType:"text",
context: this,
success : function(response) {
    console.log('RESPONSE: ' + response);
    //OPTIONAL_FUNCTION_TO_DO_SOMETHING_WITH_THE_RESPONSE(response);
},
error: function(jqXHR,textStatus,errorThrown){
    //OPTIONAL_FUNCTION_TO_DO_SOMETHING_WITH_THE_ERROR(jqXHR);
    console.log(errorThrown);
}
});
}

database.php

 if(isset($_POST['code'])){
    does_code_match($_POST['code']);
 }
 function does_code_match($code){

    connect();
            die('test');
    $sql = 'select * from emailstobeverified where email=';
    $sql .= "'" . $_SESSION['email'] . "'";
    $sql .= ' and verification_code=';
    $sql .= $code;
    $sql .= ';';
    $count = query($sql)->num_rows;

    die(strval($count));
    disconnect();
    echo strval($count);
    exit;
    //Once you've outputted, make sure nothing else happens
 }

does_code_match function executes, and the console log prints <br /> when console.log("RESPONSE" + response) is called. But I want it to print the value of $count

EDIT:

There is a problem with connect() function. If I call die('test') just before calling connect(), it returns a response! It says "hello world" in the console. If I call it directly AFTER calling connect(), it prints this:

<b>Warning</b>:  Use of undefined constant conn - assumed 'conn' (this will throw an Error in a future version of PHP) in <b>C:\xampp2\htdocs\database.php</b> on line <b>72</b><br />
<br />
<b>Warning</b>:  Use of undefined constant dbhost - assumed 'dbhost' (this will throw an Error in a future version of PHP) in <b>C:\xampp2\htdocs\database.php</b> on line <b>72</b><br />
<br />
<b>Warning</b>:  Use of undefined constant dbuser - assumed 'dbuser' (this will throw an Error in a future version of PHP) in <b>C:\xampp2\htdocs\database.php</b> on line <b>72</b><br />
<br />
<b>Warning</b>:  Use of undefined constant dbpass - assumed 'dbpass' (this will throw an Error in a future version of PHP) in <b>C:\xampp2\htdocs\database.php</b> on line <b>72</b><br />
<br />
<b>Warning</b>:  Use of undefined constant db - assumed 'db' (this will throw an Error in a future version of PHP) in <b>C:\xampp2\htdocs\database.php</b> on line <b>72</b><br />
test

and the response is <br />

Connect function:

 $dbhost = "localhost";
 $dbuser = "root";
 $dbpass = "";
 $db = "example";
 $conn;
 function connect(){
    $GLOBALS[conn] = new mysqli($GLOBALS[dbhost], $GLOBALS[dbuser], $GLOBALS[dbpass],$GLOBALS[db]) or die("Connect failed: %s
". $GLOBALS[conn] -> error);
 }
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3条回答 默认 最新

  • dongshijiao2363 2019-04-28 15:32
    关注

    When referencing array keys, you must enclose them in quotes, change your function to the following:

    function connect(){
        $GLOBALS['conn'] = new mysqli($GLOBALS['dbhost'], $GLOBALS['dbuser'], $GLOBALS['dbpass'],$GLOBALS['db']) or die("Connect failed: %s
    ". $GLOBALS['conn'] -> error);
     }
    

    Also, in general, this is a strange way of storing global variables, you should look into using the $_ENV array instead.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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