我创建了一个ajax函数来在我的数据库中输入一个表单,但我还需要携带用户ID,这样我就可以知道该表属于谁。我已经用ajax尝试过:
data: $("#add_form").serialize(), { id : <?php echo $_SESSION['user_id']; ?> },
但这不起作用,我只是收到此错误:Uncaught SyntaxError: Unexpected token ,
我如何在 Ajax 中使用表单发送我的 ID?这是我的表格:
<form action="" method="post" id="add_form" name="add_form">
<h5>Name: </h5><input type="text" name="name" class="addinput"/>
<h5>Artist: </h5><input type="text" name="artist" class="addinput"/>
<h5>Link to song: </h5><input type="text" name="url" class="addinput"/>
<h5>Lyrics: </h5><textarea name="lyrics" class="addinputtext">Here goes the lyrics...</textarea>
<input type="button" value="Add track" class="loginbtn" />
</form>
我的完整ajax调用:
// this is the id of the form
$("#add_form").submit(function() {
var url = "includes/addtrack.php"; // the script where you handle the form input.
$.ajax({
type: "POST",
url: url,
data: $("#add_form").serialize(), { id : <?php echo $_SESSION['user_id']; ?> }, // serializes the form's elements.
success: function(data)
{
alert(data); // show response from the php script.
$.ajax({ //Get the name of the artist
url: "includes/getsongs.php",
type: "POST",
data: {id: <?php echo $_SESSION["user_id"]; ?>},
success: function(data) {
//called when successful
console.log("The data is:");
console.log(data);
$(".yoursongs").html(data); //Update
},
error: function(e) {
//called when there is an error
console.log(e.message);
}
});
}
});
return false; // avoid to execute the actual submit of the form.
});
PHP 代码
<?php
include 'db_connect.php';
$stmt = $mysqli->prepare("INSERT INTO song VALUES (?)");
$stmt->bind_param('ssssi', $name, $artist, $url, $lyrics, $userid); // bind $sample to the parameter
// escape the POST data for added protection
$name = isset($_POST['name'])
? $mysqli->real_escape_string($_POST['name'])
: '';
$artist = isset($_POST['artist'])
? $mysqli->real_escape_string($_POST['artist'])
: '';
$url = isset($_POST['url'])
? $mysqli->real_escape_string($_POST['url'])
: '';
$lyrics = isset($_POST['lyrics'])
? $mysqli->real_escape_string($_POST['lyrics'])
: '';
$userid = isset($_POST['userid'])
? $mysqli->real_escape_string($_POST['userid'])
: '';
/* execute prepared statement */
$stmt->execute();
printf("%d Row inserted.\n", $stmt->affected_rows);
echo 'done';
/* close statement and connection */
$stmt->close();
?>
最佳答案
目前您的data
设置不正确,因为serialize()
会将表单值转换为查询字符串,因此您可以使用&
将您的 id
值连接到此字符串中:
data : $('#add_form').serialize() + "&id=" + <?php echo $_SESSION['user_id']; ?>,
而不是:
data: $("#add_form").serialize(), { id : <?php echo $_SESSION['user_id']; ?> },
关于javascript - 如何使用ajax发送表单和id,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23130458/