按下确认按钮后,我就可以发送数据了。
但是,当按下取消按钮时,sweetalert2显示为已成功插入数据。
后端显示一个空字符串。(在数据库表中)
当我按下取消按钮时如何验证,而不是将数据发送到后端。
function inputPass(complaintID) { // complaint id pass is ok.
swal({
text: 'Input comment message',
input: 'textarea',
showCancelButton: true,
}).then(function(sample_text) {
console.log(sample_text);
if(sample_text === '') { // problem is here.
swal({
type: 'warning',
html: 'cannot proceed without input'
});
} else {
console.log(sample_text);
$.ajax({
type: "POST",
url: "../ajax/ajax_active_deact.php?type=complaint_answered",
data: {complaintID: complaintID, sampleText: sample_text}
}).done(function (res) {
if(!res) {
swal({
type: 'error',
html: 'insert the valid text'
});
} else {
swal({
title: 'done',
text: 'all right',
type: 'success',
allowOutsideClick: false,
confirmButtonText: 'Ok'
});
}
});
}
});
}
function complaint_answered() {
include_once('../backend/ConsumerComplaint.php');
$con_complaint = new ConsumerComplaint();
$res = $con_complaint>mark_as_answered($_POST['complaintID'],$_POST['sampleText']);
echo $res;
}
function mark_as_answered($id, $comment) {
//var_dump($comment);
$val = $comment['value']; // $comment is a associative array, with the key of 'value'
$sql = "UPDATE c_consumer_complaint SET `status` = 'answered', `update` = '$val'
WHERE complaint_id = '$id' ";
$res = $this->conn->query($sql);
return $res;
}
我是开发的新手,无法解决该问题。 请任何人能告诉我我在这里做错了什么。 谢谢!
答案 0 :(得分:1)
只有在用户单击“确定”时,您才会获得result.value
,以便您可以检查是否存在一个值,如果该值为空,则显示错误消息。如果没有价值,什么也不会发生。
代码段:
swal({
text: 'Input comment message',
input: 'textarea',
showCancelButton: true,
}).then(function(result) {
if(result.value) {
$.ajax({
type: "POST",
url: "../ajax/ajax_active_deact.php?type=complaint_answered",
data: {complaintID: complaintID, sampleText: result.value}
}).done(function (res) {
if(!res) {
swal({
type: 'error',
html: 'insert the valid text'
});
} else {
swal({
title: 'done',
text: 'all right',
type: 'success',
allowOutsideClick: false,
confirmButtonText: 'Ok'
});
}
});
} else if (result.value === "") {
swal({
type: 'warning',
html: 'cannot proceed without input'
});
}
});
<script src="https://cdn.jsdelivr.net/npm/sweetalert2@7.28.7/dist/sweetalert2.all.min.js"></script>
您的课程:
在您的 php ajax代码中,您传递的$_POST['sampleText']
不是数组而是字符串,因此$comment['value']
将不包含文本。
function mark_as_answered($id, $comment) {
//var_dump($comment);
$val = $comment;
$sql = "UPDATE c_consumer_complaint SET `status` = 'answered', `update` = '$val'
WHERE complaint_id = '$id' ";
$res = $this->conn->query($sql);
return $res;
}
PS:请对SQL-Injection进行自学,以使人们无法将有害代码注入到您的SQL查询中。
答案 1 :(得分:0)
看起来像样例文本始终设置为数组。我会尝试更改if语句
if(sample_text === '') { // problem is here.
swal({
type: 'warning',
html: 'cannot proceed without input'
});
} else {
console.log(sample_text);
类似
if(sample_text['dismiss'] == 'cancel') { // problem is here.
swal({
type: 'warning',
html: 'cannot proceed without input'
});
} else {
console.log(sample_text);