Question

我想使用ajax将表单中的数据插入到数据库中。当我运行它时，它只显示index.php代码而什么都不做。我无法找出错误。所以，请帮我运行这段代码。

的index.php

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>

</head>

<body>
<input type="text" id="name" placeholder="Enter Text"> 
<button id="submit" onClick="js()" type="button"> Submit</button>

<script> 
function js() {

var name = document.getElementById("name").value ;


$.ajax({

type:'POST',
data: name,
url:"insert.php",
success:function(result){
alert(success); 
}

}); 
}

</script>
</body>



</html>

insert.php

<?php
$connection = mysqli_connect('localhost', 'root', '', 'sample');
if($_POST['name']){
$name=$_POST['name'];   
$q= "insert into test ('$name')";
$query = mysqli_query($connection, $q);
if($query){
echo 'inserted';
}
}
?>

Answer 1

如果你的表结构如下：

id int PK (Auto_increment)
name varchar(30) not null

您必须使用以下代码：

的index.php

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<script src="https://code.jquery.com/jquery-3.1.1.min.js"></script>
<title>Test</title>
</head>
<body>
<input type="text" id="name" placeholder="Enter Text"> 
<button id="submit" onClick="js()" type="button"> Submit</button>
<script> 
function js()
{
    if(document.getElementById("name").value!="")
    {
        var name = document.getElementById("name").value ;
        $.post("insert.php",{name:name},function(data){
            if(data)
            {
                alert(data);
            }
            else
            {
                alert("Cancel.");   
            }
        });
    }
    else
    {
        alert("Enter name.");
        document.getElementById("name").focus();
    }
}
</script>
</body>
</html>

insert.php

<?php
    $connection = mysqli_connect('localhost', 'root', '', 'test');
    if($_POST['name'])
    {
        $name=$_POST['name'];   
        $q= "insert into test(name) values ('$name')";
        $query = mysqli_query($connection, $q);
        if($query)
        {
            echo 'inserted';
        }
        else
        {
            echo 'no inserted';
        }
    }
?>

如果您有任何疑惑，请告诉我。

...谢谢

Answer 2

请尝试以下内容......这可能对您有所帮助..

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
<script src="https://code.jquery.com/jquery-3.1.1.min.js"></script>
</head>

<body>
<input type="text" id="name" placeholder="Enter Text"> 
<button id="submit" onClick="js()" type="button"> Submit</button>

<script> 
function js() {

var name = $("#name").val();

$.ajax({

type:'POST',
data: 'name='+name,
url:"insert.php",
success:function(result){
alert(success); 
}

}); 
}

</script>
</body>
</html>

PHP脚本如下所示。

<?php
$connection = mysqli_connect('localhost', 'root', '', 'sample');
if($_POST['name']){
$name=$_POST['name'];   
$q= "INSERT INTO test (name) VALUES ('$name')";
$query = mysqli_query($connection, $q);
if($query){
echo 'inserted';
}
}
?>

使用ajax插入MySql数据库

2 个答案: