使用mysql查詢時JSON/PHP不工作

Question

我正在嘗試使用json，出於某種原因，直到我實現任何類型的mysql時，這都可以正常工作。我從教程中偷了這個。我嘗試使用準備好的語句和基本的mysqli查詢;但是，當我這樣做時，不會返回任何數據。

這裏是jQuery的：

<script type="text/javascript"> 
$("document").ready(function(){ 
    $(".sendText").submit(function(){ 
    $(".errors").html(""); 
    $(".success").html(""); 
    var data = { 
     "action": "test" 
    }; 
    data = $(this).serialize() + "&" + $.param(data); 
    $.ajax({ 
     type: "POST", 
     dataType: "json", 
     url: "jsonpost.php", //Relative or absolute path to response.php file 
     data: data, 
     success: function(data) { 
     if(data["success"]=="yes") { 
      $(".success").html("Message Sent!"); 
      $(".formContainer").html("" + data["json"] + ""); 
     } else { 
      $(".errors").html("" + data["errors"] + ""); 
     } 
     } 
    }); 
    return false; 
    }); 
}); 
</script> 

這裏是PHP：

<?php 
    header('Content-Type: application/json'); 
    require "../database/db.php"; 

    if (is_ajax()) { 
    if (isset($_POST["action"]) && !empty($_POST["action"])) { //Checks if action value exists 
     $action = $_POST["action"]; 
     switch($action) { //Switch case for value of action 
     case "test": test_function(); break; 
     } 
    } 
    } 

    //Function to check if the request is an AJAX request 
    function is_ajax() { 
    return isset($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest'; 
    } 

    function test_function(){ 
    $return = $_POST; //to reference post 

    $phone=$return[phone]; 
    $yphone=$return[yphone]; 
    $alert=$return[alert]; 

    function generateRandomString($length) { 
     $characters = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'; 
     $randomString = ''; 
     for ($i = 0; $i < $length; $i++) { 
     $randomString .= $characters[rand(0, strlen($characters) - 1)]; 
     } 
     return $randomString; 
    } 
    $confirmCode=generateRandomString(6); 

    //validate sender phone number 
    if(empty($phone)){ 
     $c="0"; $empty="<li>Please enter their phone number!</li>"; 
    } 

    //see if content is 140 characters or less 
    if(strlen($content)>140){ 
     $c="0"; $lerror="<li>Your message must be 140 characters or less in length!</li>"; 
    } 

    /* IP Address Info */ 
    $ip=$_SERVER["REMOTE_ADDR"]; //get sender IP address 
    $ip=hash('sha256', $ip); //hash IP address 
    /* END IP Info */ 

    $etime=time()-18000; //5 hours ago from current time 
    $time=time(); //current time 

    /*This makes the code fail */ 
    $result = mysqli_query($con,"SELECT * FROM messages WHERE ip='$ip' AND phone='$phone' AND time BETWEEN $etime AND $time"); //check if ip has sent more than 5 messages to phone number in less than 5 hrs. 
    $totaltexts=mysqli_num_rows($result); 
    if($totaltexts>4){ 
     $c="0"; $exceedmessage="<li>You have sent $phone the limit of 5 messages per 5 hours. Please try again later.</li>"; 
    } 
    /* ------------------------- */ 

    if($c=="0"){ 
     $return["success"]="no"; 
     $return["errors"]="$empty $phoneerror $hooderror $yphoneerror $lerror $exceedmessage $exceedmessagea $exceedmessage2"; 
    } else { 
     $return["success"]="yes"; 
    } 

    $return["json"] = json_encode($return); 
    echo json_encode($return); 
    } 
?> 

Answer 1

我沒有看到任何地方，你建立連接$ CON。連接建立到數據庫的連接。一旦你有了，你可以查詢它。

你應該有類似下面的（很明顯，更換了正確的值）：

$con = mysqli_connect("myhost","myuser","mypassw","mydb") or die("Error " . mysqli_error($link));