傳遞變量從Javascript到PHP和背部

Question

控制檯結束回寫

Uncaught ReferenceError: provhtml is not defined

此外，當我正下方改變了我的Javascript被代碼，控制檯最終只是寫「<?= $provhtml ?>」

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

我不完全確定問題是什麼。我希望控制檯內寫入$ provhtml當用戶點擊一個名爲otherlistbut 按鈕（我也是，如果有需要是不知道d.preventDefault(); 這裏是我的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); 
?> 

這裏是我的在我的index.html

$('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

腳本在你prov.php文件

<?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 
?> 

在您的index.html

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); 
     } 
    }); 
}