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javascript - 如何在ajax成功方法()中显示引导成功消息?

转载 作者:行者123 更新时间:2023-12-01 05:28:55 25 4
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我正在使用以下 jQuery/Ajax 方法来验证表单。我将 json 数据传递到名为 add.php 的 php 文件。现在,如果 add.php 页面发现任何错误,则显示错误消息,否则显示成功消息。

现在,在ajax成功方法中,如果没有错误,我想显示引导成功消息class,否则显示错误消息

到这一行:

$('#form_result').append('<p class="alert alert-danger">'+value+'</p>');

现在我无法确定如何检查结果是否成功。

Jquery/Ajax 代码:

<script type="text/javascript">
$(document).ready(function() {
$("#add_zone").submit(function(e) {
e.preventDefault();
$.ajax({
url : 'add',
data : $(this).serialize(),
dataType : 'json',
type : 'POST',
beforeSend : function () {
$("#submit_button").val("Wait...");
},
success : function ( result ) {
$("#submit_button").val("Add New Zone");
$('#form_result').html('');
$.each( result, function( key, value ) {
if(key !== 'error') {
$('#form_result').append('<p class="alert alert-danger">'+value+'</p>');
}
});
},
});
});
});
</script>

add.php页面

if(isset($_POST['form_name']) && $_POST['form_name'] == "zone") {
if(verifyForm('zone', 'add')) {
$msg = array();
$msg['error'] = false;
$zone_name = validate_data($_POST['zone_name']);
$remark = validate_data($_POST['remark']);
$errors = array();
$check = mysqli_query($conn, "SELECT zone_name FROM zone WHERE uid ='$uid' AND zone_name = '$zone_name' ");
$num_check = mysqli_num_rows($check);

if(isset($zone_name, $remark)) {
if(empty($zone_name)) {
$msg[] = 'Zone name required';
$msg['error'] = true;
} elseif($num_check > 0 ) {
$msg[] = 'Zone name already exists, choose another name';
$msg['error'] = true;
}

if(!empty($errors)) {
$msg[] = '<div class="alert alert-danger">';
$msg[] = '<strong>OPPS! Correct the following error(s):</strong><br/>';
foreach($errors as $er) {
$msg[] = $er.'.<br/>';
$msg['error'] = true;
}
$msg[] = '</div>';
}

if(empty($errors) && $msg['error'] === false) {
$insert = mysqli_query($conn, "INSERT INTO zone (zone_name, uid, remark) VALUES('$zone_name', '$uid', '$remark') ");
if($insert) {
$msg[] = 'New zone added.';
} else {
$msg[] = "Can't add new zone.";
$msg['error'] = true;
}
}
}
echo json_encode($msg);
}
}

最佳答案

不要将输出与错误状态混合在一起:

<?php
/**
*/
if(isset($_POST['form_name']) && $_POST['form_name'] == "zone") {
if(verifyForm('zone', 'add')) {
$msg = array();
$msg['error'] = false;
$msg['body'] = [];
$zone_name = validate_data($_POST['zone_name']);
$remark = validate_data($_POST['remark']);
$errors = array();
$check = mysqli_query($conn, "SELECT zone_name FROM zone WHERE uid ='$uid' AND zone_name = '$zone_name' ");
$num_check = mysqli_num_rows($check);

if(isset($zone_name, $remark)) {
if(empty($zone_name)) {
$msg['body'][] = 'Zone name required';
$msg['error'] = true;
} elseif($num_check > 0 ) {
$msg['body'][] = 'Zone name already exists, choose another name';
$msg['error'] = true;
}

if(!empty($errors)) {
$msg['body'][] = '<div class="alert alert-danger">';
$msg['body'][] = '<strong>OPPS! Correct the following error(s):</strong><br/>';
foreach($errors as $er) {
$msg['body'][] = $er.'.<br/>';
$msg['error'] = true;
}
$msg['body'][] = '</div>';
}

if(empty($errors) && $msg['error'] === false) {
$insert = mysqli_query($conn, "INSERT INTO zone (zone_name, uid, remark) VALUES('$zone_name', '$uid', '$remark') ");
if($insert) {
$msg['body'][] = 'New zone added.';
} else {
$msg['body'][] = "Can't add new zone.";
$msg['error'] = true;
}
}
}
$msg['body'] = implode('',$msg['body']);
echo json_encode($msg);
}
}
?>
<script type="text/javascript">
$(document).ready(function() {
$("#add_zone").submit(function(e) {
e.preventDefault();
$.ajax({
url : 'add',
data : $(this).serialize(),
dataType : 'json',
type : 'POST',
beforeSend : function () {
$("#submit_button").val("Wait...");
},
success : function ( result ) {
$("#submit_button").val("Add New Zone");
if (result.error) {
$('#form_result').append('<p class="alert alert-danger">'+result.body+'</p>');
}
else {
$('#form_result') . html(result.body);
}
},
});
});
});
</script>

关于javascript - 如何在ajax成功方法()中显示引导成功消息?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38538143/

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