Question

我有一些代码如何使用json链接php。

＆＃13;

<?php

$con = mysqli_connect('localhost', 'ganesh', '5565','ganesh');
$sql = "INSERT INTO `equation`VALUES('','123','123')";
$result = mysqli_query($con, $sql);

 echo ' Updated successfully';

?>

＆＃13;

<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js"></script>

</head>
<body>
<script>

		$(document).ready(function(){
    $("button").click(function(){
        $.getJSON("insert12.php", function(result){
            $.each(result, function(result){
                $("#file").html(result);
            });
        });
    });
});
		
</script>

<input type="text" id="input"placeholder="Search..." >
<button>solve</button>

<div id="result">
  <ul></ul>
</div>
<ul id="file">
  
  
</ul>


</body>
</html>

＆＃13;

json不工作如何获取输入数据请帮帮我。例如输入（某些文本）输出（某些文本）是使用json打印的

Answer 1

在PHP代码中，您必须以JSON格式提供响应

<?php

$con = mysqli_connect('localhost', 'ganesh', '5565','ganesh');
$sql = "INSERT INTO `equation`VALUES('','123','123')";
$result = mysqli_query($con, $sql);

$response=array("type"=>"success","message"=>"Updated successfully");
ob_clean();
echo json_encode($response);

Answer 2

尝试对结果进行编码，使用json_encode($response)将数组转换为JSON格式，如@Haresh所述，并添加header('Content-Type: application/json');以避免使用echo

<?php
header('Content-Type: application/json');
$con = mysqli_connect('localhost', 'ganesh', '5565','ganesh');
$sql = "INSERT INTO `equation`VALUES('','123','123')";
$result = mysqli_query($con, $sql);

json_encode(array("type"=>"success","message"=>"Updated successfully"));

?>

json无法使用代码

2 个答案: