php - 如何在 php 中将数据库表转换为 JSON

Question

我使用了这个代码

$dbhandle = mysql_connect("localhost","root","password")
or die("Unable to connect to MySQL");

$arr;

//select a database to work with
$selected = mysql_select_db("info",$dbhandle)
or die("Could not select examples");

$result = mysql_query("SELECT * FROM students");

while($row = mysql_fetch_array($result))
{
    $arr[]=$row;
}

将表的所有数据存储到二维数组中是否正确？我如何在 Ajax 中使用这个数组作为 json 来显示 HTML 文件中的数据。

Answer 1

正如您在评论中提到的，您想要使用 ajax 和 json。

您可以执行这样的代码(使用 mysql_fetch_assoc() 而不是数组):

//set content type: text/json
header("Content-Type: text/json");

$dbhandle = mysql_connect("localhost","root","password") or die("Unable to connect to MySQL");

//select a database to work with
$selected = mysql_select_db("info",$dbhandle) or die("Could not select examples");

$result = mysql_query("SELECT * FROM students");

if(mysql_num_rows($result)>0){
    $rows=array();
    while($row = mysql_fetch_assoc($result))
    {
        $rows[]=$row;
    }
    echo json_encode(array("success"=>$rows));
}else{
    echo json_encode(array("error"=>"No records found."));
}
exit;

对于 ajax(假设 jQuery.get() AJAX )

$.get("script.php", function(data){
    if(data.success){
        //process each row
        data.success.forEach(function(row){
            console.log(row); //access student row here, simply like row.student_name, row.city etc.
        });
    }else{
        alert(data.error);
    }
});

安全说明:

1. mysql_* functions are deprecated and will be removed from support, so you should avoid using it .
2. 使用mysqli_*或PDO用于数据库查询。