php 我试图获得JSON数据成功响应，但我得到了错误

lf3rwulv 于 2023-04-28 发布在 PHP

关注(0)|答案(1)|浏览(153)

我需要为每个成功发送的消息获取JSON响应。我可以成功发送短信，但我无法从send_sms收到成功消息。PHP

<!-- index.php --->
<div id="responce"></div> <!-- respond sms sending -->
<form method="POST" onclick="return false">
<div class="form-group">
    <label>Your Message</label>
<textarea class="form-control" rows="4" cols="100" id="message"
name="message" maxlength="70" required></textarea>
</div>
<button class="btn btn-primary" type="submit" name="submit"
onmousedown="start_sendig();" class="btn">Send</button>
</form>

//sendsms_button.js

function start_sendig() {
  $("#responce").html("Please wait sending on process");
  var message = $("#message").val();
  var path_uri = "sms/send_sms.php";

  $.ajax({
    type: "POST",
    url: path_uri,
    data: {
      message: message,
    },
    success: function (data) {
      var json = $.parseJSON(data);

      if (json.response == "success") {
        $("#responce").html(
          "Message Sent Successfully to " + json.current + " !!"
        );
      } else {
        $("#responce").html("Error to Sent " + json.current + " !!");
      }
    },
  });
}

AJAX 显示此错误

caught SyntaxError: Unexpected end of JSON input
    at Function.parse [as parseJSON] (<anonymous>)
    at Object.success (sendsms_button.js:13:20)
    at i (jquery-3.2.1.min.js:2:28017)
    at Object.fireWith [as resolveWith] (jquery-3.2.1.min.js:2:28783)
    at A (jquery-3.2.1.min.js:4:14035)
    at XMLHttpRequest.<anonymous> (jquery-3.2.1.min.js:4:16323)

<?php
//send_sms.php
   
include ("vendor/autoload.php");
include ("../config.php");

//insert record
$msg_sms = $_POST['message'];
mysqli_query ($con, "insert into sent_sms (message) values ('$msg_sms')");
if (isset($_POST['submit'])){
//select numbers
$result = mysqli_query($con, "SELECT mphone FROM members");
while ($row = mysqli_fetch_array($result)){
$numbers = $row['mphone'];

    foreach (explode("+", $numbers) as $phone) {
    $basic  = new \Vonage\Client\Credentials\Basic("fxxxxx", "xxxxxxxxxxxxx");
    $client = new \Vonage\Client($basic);

    $response = $client->sms()->send(
    new \Vonage\SMS\Message\SMS($phone, 'Hotel', $msg_sms)
    );

    $message = $response->current();

        if ($message->getStatus() == 0) {
            $data = array(
                    "response" => "success",
                    "current" => '+' . $phone
                );

                echo json_encode($data);
        } else {
        echo "<script>alert('The message failed with status: " . $message->getStatus() . "');</script>";
        }
    }
}

}
   
 
?>

我的目标是显示每个成功消息，如发送到+125332626成功

php

来源：https://stackoverflow.com/questions/76082484/am-trying-to-get-json-data-success-response-but-i-got-error

1条答案

按热度按时间

62lalag41#

在循环中回显echo json_encode($data);会产生问题。.将其保存在数组中，然后在循环后回显整个数组
返回<script>不是一个好主意。.好的方法是将错误作为编码JSON返回，然后 AJAX 成功函数上检查它
-首先我们假设你的php代码和mysql连接正常。.
-第二个在 AJAX 中，你需要添加dataType : "json"。.
-第三个还需要检查isset($_POST['message'])，而不是isset($_POST['submit'])

function start_sendig() {
  $("#responce").html("Please wait sending on process");
  var message = $("#message").val();
  var path_uri = "sms/send_sms.php";

  $.ajax({
    type: "POST",
    url: path_uri,
    dataType : "json",  // <<<<<<<<<<<<<<<
    data: {
      message: message,
    },
    success: function (data) {
      if(data.error){   // <<<<<<<<<<<<<<<
         console.log(data.error);
      }else{
         console.log(data);
      }
    },
  });
}

在php中

<?php
   include ("vendor/autoload.php");
   include ("../config.php");
   if (isset($_POST['message'])){
     //insert record
     $msg_sms = $_POST['message'];
     mysqli_query ($con, "insert into sent_sms (message) values ('$msg_sms')");
    //select numbers
    $result = mysqli_query($con, "SELECT mphone FROM members");
    $returned_JSON = []; //<<<<<<<<<<<<<<<<<<<<<<<
    while ($row = mysqli_fetch_array($result)){
    $numbers = $row['mphone'];
        foreach (explode("+", $numbers) as $phone) {
        $basic  = new \Vonage\Client\Credentials\Basic("fxxxxx", "xxxxxxxxxxxxx");
        $client = new \Vonage\Client($basic);
    
        $response = $client->sms()->send(
        new \Vonage\SMS\Message\SMS($phone, 'Hotel', $msg_sms)
        );
    
        $message = $response->current();
    
            if ($message->getStatus() == 0) {
                $data = array(
                        "response" => "success",
                        "current" => '+' . $phone
                    );
    
                    $returned_JSON[] = $data; //<<<<<<<<<<<<<<<<<<<<<<<
            } else {
                $returned_JSON['error'] = $message->getStatus(); //<<<<<<<<<
            }
        }
    }
    echo json_encode($returned_JSON); //<<<<<<<<<<<<<<<<<<<<<
    }
       
     
    ?>

**注意：**使用 AJAX 时有很多原因会出错。我给了你一些代码的指导，让你知道如何思考和编码它。.但是你必须一步一步地检查 AJAX /php连接，以确切地知道什么是工作，什么是不.

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php 我试图获得JSON数据成功响应，但我得到了错误

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