Question

控制台最终回写

未捕获的ReferenceError：未定义provhtml

此外，当我将我的Javascript更改为下面的代码时，控制台最终只会写＆＃34; <?= $provhtml ?>＆＃34;

var provnum = "<?= $provhtml ?>";
console.log(provnum);

我不确定问题是什么。当用户点击名为otherlist的按钮时，我希望$ provhtml写在控制台内（我也不确定是否需要d.preventDefault(); 这是我的prov.php

<?php    
$phonenum = $_POST["phonenum"];
$provhtml = file_get_contents('http://api.data24-7.com/v/2.0?user=USERNAME&pass=PASSWORD&api=C&p1=1' . $phonenum);   
 ?>

以下是我的index.html中的脚本

   $('button[name="otherlistbut"]').click(function(d) {
        lookupProvider(d);

    });
});

/************ FUNCTIONS ******************/

function lookupProvider(d) {
    d.preventDefault();
    var phonenum = $('input[name="phonenum"]').val();
    $.ajax({
        type: 'POST',
        data: {
            phonenum: phonenum
        },
        url: 'prov.php',
        success: function(data) {
            var provnum = $(provhtml);
            console.log(provnum);
        },
        error: function(xhr, err) {
            console.log("readyState: " + xhr.readyState + "\nstatus: " + xhr.status);
            console.log("responseText: " + xhr.responseText);
        }
    });
}

Answer 1

在prov.php文件中

<?php    
$phonenum = $_POST["phonenum"];
$provhtml = file_get_contents('http://api.data24-7.com/v/2.0?user=USERNAME&pass=PASSWORD&api=C&p1=1' . $phonenum);   
echo json_encode( array( "phonenum" => $phonenum,"provhtml"=>$provhtml)); //You should be returning the values to your ajax function
 ?>

在index.html中

function lookupProvider(d) {
    d.preventDefault();
    var phonenum = $('input[name="phonenum"]').val();
    $.ajax({
        type: 'POST',
        data: {
            phonenum: phonenum
        },
        url: 'prov.php',
        success: function(data) { //Receives the data from the php code
            var provnum = data.phonenum;
            console.log(provnum);
        },
        error: function(xhr, err) {
            console.log("readyState: " + xhr.readyState + "\nstatus: " + xhr.status);
            console.log("responseText: " + xhr.responseText);
        }
    });
}

将变量从Javascript传递到PHP并返回

1 个答案: