Php 如果数据库中已经存在值,如何显示错误消息?
我有一个页面,用户可以通过ajax创建主题。在创建主题之前,我会检查它是否已经存在于数据库中。如果数据库中存在该主题,则在继续提交新主题之前,我希望通过div框显示错误消息。一切正常。但我不知道如何使错误消息仅在主题已存在于数据库中时显示 这是包含我的表单和ajax代码的页面:Php 如果数据库中已经存在值,如何显示错误消息?,php,jquery,mysql,ajax,Php,Jquery,Mysql,Ajax,我有一个页面,用户可以通过ajax创建主题。在创建主题之前,我会检查它是否已经存在于数据库中。如果数据库中存在该主题,则在继续提交新主题之前,我希望通过div框显示错误消息。一切正常。但我不知道如何使错误消息仅在主题已存在于数据库中时显示 这是包含我的表单和ajax代码的页面: <script> function cancelTopicCreation() { $(".create-topic-cancel-button, .close_blue_modal"
<script>
function cancelTopicCreation() {
$(".create-topic-cancel-button, .close_blue_modal").click(function () {
$('#id-create-topic-picture-preview').attr('src','<?php echo BASE_URL; ?>content/pictures/layout/default_group_icon.png');
});
}
function close_blue_modal() {
$('#blue_modal_mask').fadeOut(144);
$('.blue_modal_window').fadeOut(144);
$('.blue_modal_window_two').fadeOut(144);
$('.blue_modal_window_three').fadeOut(144);
$('.blue_modal_window_five').fadeOut(144);
}
function createTopic() {
/*----------------submit title of new topic----------------*/
$(document).ready(function() {
var topicTitle = $(".create-topic-input-box").val();/* get the value of the topic title input and put it in a variable */
var dataString = 'topicTitle='+ topicTitle;/* returns successful data submission message when the entered information is stored in database. */
if(topicTitle == '') {/* if there is no title entered then */
alert("Please Fill All Fields");
} else {
$.ajax({
type: "POST",
url: site_url+'resources/create-topic.php',
data: dataString,
cache: false,
success: function(result) {/* if the ajax submit is successful then */
$('#createTopicForm').each(function() {
this.reset();
});
close_blue_modal();
$('#id-create-topic-picture-preview').attr('src','<?php echo BASE_URL; ?>content/pictures/layout/default_group_icon.png');
$("#results").html(''); //append data into #results element
load_contents(track_page); //initial content load
}
});
}
return false;
});
/*----------------upload topic picture----------------*/
$(document).ready(function() {
var input = document.getElementById("id-create-topic-picture-input");
file = input.files[0];
if (file != undefined){
formData= new FormData();
if(!!file.type.match(/image.*/)){
formData.append("nameTopicPicture", file);
$.ajax({
url: site_url+'resources/upload-topic-icon.php',
type: "POST",
data: formData,
processData: false,
contentType: false,
success: function(data) {/* you can do something if the picture upload is successful here */}
});
} else {
alert('Not a valid image!');
}
} else{/* if there is no image selected then */}
});
}
</script>
<div id="blue_modal_mask" class="close_blue_modal"></div><!--#blue_modal_mask .close_blue_modal-->
<div id="blue_modal_four" class="blue_modal_window_two">
<div onclick="return cancelTopicCreation();" class="close_blue_modal icon-x-mark"></div><!--.close_white_modal-->
<div class="create-topic-label">Create topic</div><!--.create-topic-label-->
<div class="create-topic-error-message">That topic already exists. Please choose a different topic name.</div><!--.create-topic-label-->
<div class="create-topic-description"><div class="icon-book create-topic-description-icon"></div>Topics are used to organize content on the website.</div><!--.create-topic-label-->
<form id="createTopicForm" action="<?php echo BASE_URL?>resources/upload_topic_icon.php" method="post" enctype="multipart/form-data">
<div class="create-topic-input-wrap">
<label for='id-create-topic-input-box'><div class='create-topic-name-label'>Topic name</div></label>
<input class="create-topic-input-box" id="id-create-topic-input-box" type="text" name="topicTitle" placeholder="Enter a topic name" autocomplete="off"><!--input.index_sign_up_input-->
<div class="create-topic-autocomplete-result-box">
<div class="create-topic-autocomplete-result-list nice-scroll">
<div class="create-topic-autocomplete-result-item">
<img class="create-topic-autocomplete-result-item-picture" src="content/pictures/layout/default_group_icon.png" alt="">
<div class="create-topic-autocomplete-result-item-content">
<div class="create-topic-autocomplete-result-item-title">The title of the result goes here</div>
<div class="create-topic-autocomplete-result-item-details">The details of the result goes here</div>
</div><!--create-topic-autocomplete-result-item-content-->
</div><!--create-topic-autocomplete-result-item-->
<div class="create-topic-autocomplete-result-item">
<img class="create-topic-autocomplete-result-item-picture" src="content/pictures/layout/default_group_icon.png" alt="">
<div class="create-topic-autocomplete-result-item-content">
<div class="create-topic-autocomplete-result-item-title">The name of the result goes here</div>
<div class="create-topic-autocomplete-result-item-details">The details of the result goes here</div>
</div><!--create-topic-autocomplete-result-item-content-->
</div><!--create-topic-autocomplete-result-item-->
<div class="create-topic-autocomplete-result-item">
<img class="create-topic-autocomplete-result-item-picture" src="content/pictures/layout/default_group_icon.png" alt="">
<div class="create-topic-autocomplete-result-item-content">
<div class="create-topic-autocomplete-result-item-title">The name of the result goes here</div>
<div class="create-topic-autocomplete-result-item-details">The details of the result goes here</div>
</div><!--create-topic-autocomplete-result-item-content-->
</div><!--create-topic-autocomplete-result-item-->
<div class="create-topic-autocomplete-result-item-empty">Nothing found</div>
</div><!--create-topic-autocomplete-result-list-->
</div><!--create-topic-autocomplete-result-box-->
</div>
<div class="create-topic-picture-preview-wrap">
<img id="id-create-topic-picture-preview" src="<?php echo BASE_URL; ?>content/pictures/layout/default_group_icon.png" class="create-topic-picture-preview">
</div>
<div id="create-topic-footer">
<label for="id-create-topic-picture-input" >
<div class="icon-camera create-topic-camera-icon"></div>
</label>
<input name="nameTopicPicture" accept="image/*" type="file" class="create-topic-picture-input" id="id-create-topic-picture-input">
<a onclick="return createTopic();" class="create-topic-button">Create</a>
<a onclick="return cancelTopicCreation();" class="create-topic-cancel-button cancel_blue_modal">Cancel</a>
</div><!--#new_post_options-->
</form>
</div><!--.blue_modal_one .blue_modal_window-->
我怎样才能告诉javascript这个名称的主题已经存在。如果该主题已经存在于数据库中,那么我希望ajax停止并显示错误弹出div。请帮助。我尝试使用json传递一个字符串,表示数据库中已经存在该主题。但是没有成功。我曾尝试将字符串从php传递到查询,但问题是,多次尝试都不会改变字符串。因此每次都会显示错误。您可以让PHP发送脚本,而不是简单的字符串 类似的方法也可以:
$query = mysqli_query($databaseConnection, "SELECT columnTopicTitle FROM tableTopics WHERE columnTopicTitle='".$topicTitle."'");
if (mysqli_num_rows($query) != 0) {
//echo "Username already exists";
?>
<script>
alert("Username already exists");
</script>
<?php
} else { ....
在Ajaxsuccess
回调中:
success: function(result) {
// Rest of your code...
// Add this for the script to be sent.
$("#someEmptyDiv").html(result);
------编辑
好的。。。您必须了解可能的结果:
1-有关用户名的错误消息已存在
2-一个JSON 这里有很多方法可以发挥创意,这取决于你想做什么:
$query = mysqli_query($databaseConnection, "SELECT columnTopicTitle FROM tableTopics WHERE columnTopicTitle='".$topicTitle."'");
if (mysqli_num_rows($query) != 0) {
//echo "Username already exists";
?>
<script>
$("#myHiddenDiv").html("Username already exists").fadeIn(600);
</script>
<?php
} else {
$advert = array( // I assume this works.
'ajax' => 'Hello world!',
'advert' => $row['adverts'],
);
?>
<script>
close_blue_modal();
$("#myHiddenDiv").html("Username created").fadeIn(600);
console.log( " <?php echo json_encode($advert); ?> " );
</script>
<?php
//echo json_encode($advert);
}
这里有更多的评论,您的php在错误时回显了一个文本字符串
用户名已经存在
,但在底部也有回显
ajson_encode()
。您的错误消息将错误地形成json。它会写Username已经存在{“ajax”:“Hello world”,“advert”:“whatever”}
,如果出现错误则无效。好的,如果Username已经存在,我需要用js显示一个div。我该怎么做?我应该使用ajax返回变量还是json?我应该怎么做?@RomeoHennessy使用json只是因为你在jquery代码中循环了一个json。问题是我的json返回的是未定义的。我不知道如何正确使用JSON。@RomeoHennessy您将数据库数据存储在$row
的哪里??你错过了。为什么我的javascript不能从php页面工作?它没有任何作用。我已经试过了…它不起作用“PHP。。。它是由它发送的。。。等一下,我将添加一些内容您必须将ajax结果放入div
。如果是脚本,它将执行。把它藏起来。。。因此,如果它不是脚本,它将不会显示。好的。另一件事是,在ajax调用成功后,我关闭了允许人们提交新主题的框。但是,如果主题已经存在,那么我想保持该框打开,直到输入有效的主题。嗯。。。可以因此,明确地说,您需要询问Ajax用户名是否已经存在。看起来您只有2个可能的答案1-错误消息2-另一条消息。。。(你的Json东西我没看)。因此,在成功
中,您可以检查结果的内容以做出决策
success: function(result) {
// Rest of your code...
// Add this for the script to be sent.
$("#someEmptyDiv").html(result);
$query = mysqli_query($databaseConnection, "SELECT columnTopicTitle FROM tableTopics WHERE columnTopicTitle='".$topicTitle."'");
if (mysqli_num_rows($query) != 0) {
//echo "Username already exists";
?>
<script>
$("#myHiddenDiv").html("Username already exists").fadeIn(600);
</script>
<?php
} else {
$advert = array( // I assume this works.
'ajax' => 'Hello world!',
'advert' => $row['adverts'],
);
?>
<script>
close_blue_modal();
$("#myHiddenDiv").html("Username created").fadeIn(600);
console.log( " <?php echo json_encode($advert); ?> " );
</script>
<?php
//echo json_encode($advert);
}
success: function(result) {
$("#someEmptyDiv").html(result);
}