我有一种形式,我使用ajax提交数据。但是firebug给了我错误" ReferenceError:editUser未定义"。表单是一个模态,我正在使用表单来编辑用户信息。 请问有什么问题? 我的表格是:
<script type="text/javascript">
function editUser(ref) {
var user_id=ref;
var password= document.getElementById("password"+ref).value;
var userName= document.getElementById("userName"+ref).value;
var firstName= document.getElementById("firstName"+ref).value;
var lastName= document.getElementById("lastName"+ref).value;
var role= document.getElementById("role"+ref).value;
var resp;
if (window.XMLHttpRequest) {
resp = new XMLHttpRequest();
xmlhttp = new XMLHttpRequest();
} else if (window.ActiveXObject) {
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
var data = "user_id="+user_id+"&password="+password+"&userName="+userName+"&firstName="+firstName+"&lastName="lastName+"&role="+role
xmlhttp.open("POST",
"update_user.php");
xmlhttp.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
xmlhttp.send(data);
xmlhttp.onreadystatechange =
function display_data() {
if (xmlhttp.readyState == 4 && xmlhttp.status==200) {
alert(resp.responseText);
location.reload(true);
}
}
}
</script>
echo '<div class="modal fade" id="users'.$ref.'" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true">';
echo '<div class="modal-dialog">';
echo '<div class="modal-content">';
echo '<div class="modal-header">';
echo '<button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">×</span></button>';
echo '<h4 class="modal-title">Edit User Information</h4>';
echo' </div>';
echo '<div class="modal-body">';
echo '<form method="post" action="users.php" id="edit">';
echo '<label class="control-label col-md-3" >Username:</label><input class="form-control" name="userName" id="userName" style="width:310px" type="text" value="'.$row['userName'].'" required><br>';
echo '<label class="control-label col-md-3" >Password:</label><input class="form-control" name="password" id="password" style="width:310px" type="text" value="'.$row['password'].'"required><br>';
echo '<label class="control-label col-md-3" >First Name:</label><input class="form-control" name="firstName" id="firstName" style="width:310px" type="text" value="'.$row['firstName'].'" required><br>';
echo '<label class="control-label col-md-3" >Last Name:</label><input class="form-control" name="lastName" id="lastName" style="width:310px" type="text" value="'.$row['lastName'].'" required><br>';
echo '<label class="control-label col-md-3" >Role:</label>';
echo ' <select class="form-control" style="width:310px" name="role" id="role">';
echo '<option value="'.$row['role'].'">'.$row['role'].'</option>';
echo '<option value="Administrator">Administrator</option>';
echo '<option value="Admissions">Admissions Officer</option>';
echo '<option value="Finance">Finance Officer</option>';
echo '<option value="Lecturer">Lecturer</option>';
echo '<option value="Timetable">Time Tabling Officer</option>';
echo ' <option value="Librarian">Librarian</option>';
echo '<option value="Hostelier">Accomodation Officer</option>';
echo '<option value="Election">Election Officer</option>';
echo '<option value="Health">Health Officer</option>';
echo '</select>';
echo'<br>';
echo'</form>';
echo'</div>';
echo'<div class="modal-footer">';
echo'<button type="button" class="btn btn-default btn-danger" data-dismiss="modal"><i class="glyphicon glyphicon-remove glyphicon-large"></i>Cancel</button>';
echo'<button type="button" class="btn btn-default btn-success" data-dismiss="modal" onclick="editUser('.$ref.')">Save</button>';
echo'</div>';
echo'</div>';
echo'</div>';
echo '</div>';
echo '</div>';
echo "<script> $('#users$ref').modal(show)</script>";
update_user.php:
<?php include('includes/conn.php');?>
<?php
$id = intval($_POST['user_id']);
$userName = $_POST['userName'];
$password = $_POST['password'];
$firstName = $_POST['firstName'];
$lastName = $_POST['lastName'];
$role = $_POST['role'];
$sql="UPDATE users SET userName='$userName',password='$password',firstName='$firstName',lastName='$lastName',role='$role' WHERE id='$id' ";
$query=mysqli_query($query) or die(mysqli_error($conn));
if($query)
{
echo "Operation was successful";
}
else
{
echo "An error occured.Retry";
}
?>
答案 0 :(得分:0)
我使用了类似的功能发布到聊天中,我在表格页面上使用了这段代码:
JQuery:<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3/jquery.min.js"></script>
脚本:
<script type="text/javascript">$(document).ready(function(){ //IF THE USER SUBMIT THE FORM
$("#submitform").click(function(){
var input_1 = $("#input_1").val();
$.post("/submit.php", {text: input_1});
$("#input_1").attr("value","");
return false;
});
});</script>
Submit.php:<?php $text = $_POST['input_1']; //Your code
?>
这里不是真的用来回答问题,所以关于如何在将来提供更好帮助的任何指导都会有所帮助:)