使用jQuery將字符串從php發送到js文件

Question

我需要從php文件中獲取字符串有人知道我該怎麼做？

My.php即寫入和讀取一個文件的contenute使用該梅索德

<?php 
$dir = '/var/www/devData/test'; 

// create new directory with 777 permissions if it does not exist yet 
// owner will be the user/group the PHP script is run under 
if (!file_exists($dir)) { 
    mkdir ($dir, 0777); 
} 
if (isset($_POST['data'])) { 

     $stringData = $_POST['data']; 
     $file = "/var/www/devData/test/ciao.txt"; 
     $fh = fopen($file, 'r+') or die("can't open file"); 
     fwrite($fh, $stringData); 
     $theData = fread($fh, filesize($myFile)); //this is the string that i have to pass 
     fclose($fh); 

     return $theData; 

} 


?> 

my.js SCRPT得到從我的PHP文件retrive的strng

function addLabelCustom_options() { 
    var select = document.getElementById('label_custom'); 

$.get("JS/foo.php", function(result){ alert(result) }, "json"); 


/* $.ajax({ 
    url:'/var/www/devData/test/ciao.txt', 
    success: function (data){ 
    //parse ur data 
    //you can split into lines using data.split('\n') 
    //use regex functions to effectivley parse 
    var label_parsed = data.splitCSV(); 
    select.options[0] = new Option("-- Select Label --",0); 
    var label_sort = new Array(); 

    for (var i=0; i<label_parsed.length-1; i++) 
     label_sort.push(label_parsed[i][1]); 

    label_sort.sort(); 
    for (var j=0; j<label_sort.length-1; j++) 
     select.options[j+1] = new Option(label_sort[j],j+1); 

} 
}); 

*/ 
} 

我只需要得到字符串的含義$ theData 看起來好像我不回收任何字符串....

Answer 1

<?php 
$dir = '/var/www/devData/test'; 

// create new directory with 777 permissions if it does not exist yet 
// owner will be the user/group the PHP script is run under 
if (!file_exists($dir)) { 
    mkdir ($dir, 0777); 
} 
if (isset($_POST['data'])) { 

     $stringData = $_POST['data']; 
     $file = "/var/www/devData/test/ciao.txt"; 
     $fh = fopen($file, 'r+') or die("can't open file"); 
     fwrite($fh, $stringData); 
     $theData = fread($fh, filesize($myFile)); //this is the string that i have to pass 
     fclose($fh); 

     echo $theData; // <-- ECHO 

} 


?> 

你必須輸出數據，而不是返回它。那就是迴應它。如果字符串尚未JSON編碼，使用echo json_encode($theData);

Answer 2

呼應PHP函數的內容之後（回聲$海圖;）試試下面的jQuery代碼

$.ajax({ 
url:JS/foo.php, 
success:function(result){ alert(result) }, "json"), 
error:function(alert('Not Completed')) 
}; 

而且還要檢查控制檯響應檢查任何errors.You可以查看火狐的F12控制檯

Answer 3

我解決了這個問題，每次都是愚蠢的事情，每次我明白我真的很愚蠢，但我後解決方案。

冷杉PHP文件用於在文件

<?php 
$dir = '/var/www/devData/test'; 

// create new directory with 777 permissions if it does not exist yet 
// owner will be the user/group the PHP script is run under 
if (!file_exists($dir)) { 
    mkdir ($dir, 0777); 
} 
if (isset($_POST['data'])) { 

     $stringData = $_POST['data']; 
     $file = "/var/www/devData/test/ciao.txt"; 
     $fh = fopen($file, 'r+') or die("can't open file"); 
     fwrite($fh, $stringData); 
     $theData = fread($fh, filesize($file)); 
     fclose($fh); 

     echo json_encode($theData); 
} 


?> 

創建一個目錄和存儲讀我創建另一個文件PHP

<?php 


     $file = "/var/www/devData/test/ciao.txt"; 
      $fh = fopen($file, 'r') or die("can't open file"); 
     $theData = fread($fh, filesize($file));<--- was the error before i used Myfile but that variable wasn't declare and was empty :) lol 
     fclose($fh); 

     echo json_encode($theData); 


?> 

感謝大家的contenute！