ajax失敗並返回一個網頁，而不是json

-1

我想從JavaScript腳本中導入一個json數組，但每次請求失敗時，xhr.responseText都會返回關聯的html模板的模板的html代碼到它。我正在使用xampp。ajax失敗並返回一個網頁，而不是json

PHP代碼：

<?php 
// This script is meant to display the data from Insight sensors DB 

include('includes/motebymote.html'); 

if ($_SERVER['REQUEST_METHOD'] == 'POST'){//Use the post method as a request method to the server 

    require('mysqli_connect.php');//Give information on which database to use 


    if(!empty($_COOKIE["mote_ID"]) || !empty($_COOKIE["std"]) || !empty($_COOKIE["end"])){ 

     //Setting variables 
     $moteID=$_COOKIE["mote_ID"]; 
     $start=$_COOKIE["std"]; 
     $end=$_COOKIE["end"]; 

     if(preg_match("/^[0-9]{4}-[0-9]{2}-[0-9]{2}$/", $start) || preg_match("/^[0-9]{4}-[0-9]{2}-[0-9]{2}$/", $end)) { 

      if(isset($moteID) && isset($start) && isset($end)){ 

       //Building the query 
       $q="SELECT date_time, modality, value FROM sensor_data WHERE mote_id = $moteID and date_time BETWEEN '$start' and '$end'"; 

       $r = @mysqli_query ($dbc, $q);//Doing the request 

       $numRow = mysqli_num_rows($r); 


       if ($r && $numRow>0){ 
        //echo "<p>There are $numRow data points associated to mote$moteID</p><br>"; 

        //echo"<table align ='center' width ='70%' cellpadding='4'><tr><th>Date&Time</th><th> Sensor Type</th><th> Value</th></tr>"; 

        $moteArray = array(); 
        while($data = mysqli_fetch_array($r, MYSQLI_BOTH)){ 
         //echo "<tr align ='center'><td>" .$data['date_time']. "</td><td>" .$data['modality']. "</td><td>" .$data['value']. "</td></tr>"; 
         $moteArray[] = $data;      
        } 
        //echo"</table>"; 
        $JSONMoteArray = json_encode($moteArray); 
        echo $JSONMoteArray; 
       } 
       else if ($numRow==0){ 
        echo "<h1>Sorry there are no data available</h1> 
        <p>There are no data available for mote$moteID</p>"; 
       } 
       else{//Give an error message if the query fails 
        echo '<h1>System Error</h1> 
        <p class="error">We could not record your order due to a system error. We apologize for any inconvenience.</p>'; 
       } 
      } 
     } 
     else { 
      echo "<h1> Wrong date format</h1> 
      <p>Please input the starting date in a yyyy-mm-dd format</p>"; 
     } 
    } 
    else{ 
     echo "<h1> Wrong input</h1> 
     <p> Please make sure that the mote is properly selected and that the start and end dates are entered in yyyy-mm-dd format</p>"; 
    } 
    mysqli_close($dbc); //close the database 
} 

exit(); //exits the script 
?>

<!DOCTYPE html> 
<html lang="eng"> 
    <head> 
    <meta charset="utf-8"> 
    <title> Motes Data</title> 
    <script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script> 
    <script type="text/javascript" src="js/mote.js" async></script> 
    <script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script> 
    </head> 
    <body><br> 
     <form action="motebymote.php" method="post"> 
      <p>Please input a mote number: <input type="text" name="moteNo" id="moteNo"></p> 
      <p> Please input a starting date (yyyy-mm-dd):<input type="text" name="stDate" id="stDate"></p> 
      <p> Please input an end date (yyyy-mm-dd):<input type="text" name="endDate" id="endDate"></p> 
      <p><input type="submit" name="submit" value="submit" id="submit"/></p> 
      <p><input type="submit" name="display" value="display" id="display"/></p> 
      <div id="columnchart_material" style="width: 800px; height: 500px;"></div> 
     </form> 
    </body> 
</html>

var but = document.getElementById("submit"); //Looking for the submit button 

if(but){           //if the button is clicked activate the event listener 
    but.addEventListener("click", moteIDDate); 
} 

//This function goes to fetch the different values and store them with cookies //to allow motebymote.php to use them 
function moteIDDate(){ 

    var moteID = $('#moteNo').val(); 
    document.cookie="mote_ID = " + moteID + ";time()+3600"; 

    var start = $("#stDate").val(); 
    document.cookie = "std="+start+";time() 3600"; 

    var end = $("#endDate").val(); 
    document.cookie= "end="+ end+";time() 3600"; 
}; 

var jsonData = $.ajax({ 
    url: "http://localhost/Insight/mote2/motebymote.php", 
    type:"GET", 
    cache: false, 
    dataType: "json", 
    success: function(xhr){ 
     console.log(xhr.responseText); 
    }, 
    error: function(xhr){ 
    console.log(xhr.responseText); 
    } 
});

來源

2017-04-25 Alex Acquier

許多問題，以及其中：** 1）**添加'error_reporting（E_ALL）; ini_set（'display_errors'，1）;'在你的PHP頁面之上（因此我們知道是否有錯誤）** 2）**爲什麼使用PHP響應模板？（query - > result - > json - > response - > json use - > tpl it，no？）** 3）** Ajax GET到PHP POST？ ** 4）* *請*擺脫'@' – OldPadawan

只在頁面中包含你的html，如果你沒有發佈到頁面.... +下面的答案。 –

我可以從PHP頁面回顯JSON數組我認爲PHP代碼是好的，但我是編程中的新東西你是什麼意思通過查詢 - >結果 - > JSON - >響應 - > JSON使用 - > TPL它，不是？ oldPadawan –

PHP代碼中有「如果...... == POST」你的JavaScript Ajax調用，雖然發送「GET」

來源

2017-04-25 13:26:06 RC1234

如果我嘗試引入POST，它將返回html代碼加上我想要的json數組，但是作爲字符串，它也出現在錯誤函數的控制檯日誌中 –

ajax失敗並返回一個網頁，而不是json

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