Javascript 更改datatable中的警告消息
我有datatable，我有使用两个组合框的高级搜索，我的问题是如果我搜索一些数据，但数据不存在，它会显示如下警告DataTables警告：table id=myTable-无效JSON响应。有关此错误的详细信息，请参阅。现在我想改变警告，就像没有工作一样。请帮帮我。我只是个学生，对不起我的英语 这是我在my search.php中的代码Javascript 更改datatable中的警告消息,javascript,php,jquery,ajax,datatables,Javascript,Php,Jquery,Ajax,Datatables,我有datatable，我有使用两个组合框的高级搜索，我的问题是如果我搜索一些数据，但数据不存在，它会显示如下警告DataTables警告：table id=myTable-无效JSON响应。有关此错误的详细信息，请参阅。现在我想改变警告，就像没有工作一样。请帮帮我。我只是个学生，对不起我的英语 这是我在my search.php中的代码 <section> <div class="container"> <div class="row">
<section>
<div class="container">
<div class="row">
<div class="col-md-12">
<form id="myForm" class="form-inline">
<div class="form-group">
<label>Experience</label>
<select id="experience" class="form-control" name="exp">
<option disabled selected>Select Experience</option>
<?php
$sql = "SELECT DISTINCT(experience) FROM job_post WHERE experience is not null";
$result=$conn->query($sql);
if ($result->num_rows>0) {
while ($row = $result->fetch_assoc()) {
echo "<option value='".$row['experience']."'>".$row['experience']."</option>";
}
}
?>
</select>
</div>
<div class="form-group">
<label>Qualification</label>
<select id="qualification" class="form-control" name="qua">
<option disabled selected>Qualification</option>
<?php
$sql = "SELECT DISTINCT(qualification) FROM job_post WHERE qualification is not null";
$result=$conn->query($sql);
if ($result->num_rows>0) {
while ($row = $result->fetch_assoc()) {
echo "<option value='".$row['qualification']."'>".$row['qualification']."</option>";
}
}
?>
</select>
</div>
<button id="" class="btn btn-success">Search</button>
</form>
</div>
</div>
<div class="row" style="margin-top: 5%;">
<div class="table-responsive">
<table id="myTable" class="table" style="width: 1190px;">
<thead>
<th>Job Name</th>
<th>Job Description</th>
<th>Minimum Salary</th>
<th>Maximum Salary</th>
<th>Experience</th>
<th>Qualification</th>
</thead>
<tbody>
</tbody>
</table>
</div>
</div>
</div>
</section>
<script src="js/jquery-3.3.1.min.js"></script>
<script src="//cdn.datatables.net/1.10.19/js/jquery.dataTables.min.js"></script>
<script type="text/javascript">
$(function(){
var oTable = $('#myTable').DataTable({
"autoWidth" : false,
"ajax" : {
"url" : "refresh_job_search.php",
"dataSrc": "",
"data" : function(d){
d.experience = $("#experience").val();
d.qualification = $("#qualification").val();
}
}
});
$("#myForm").on("submit", function(e){
e.preventDefault();
oTable.ajax.reload(null, false);
});
});
</script>
<?php
session_start();
require_once("db.php");
$sql = "SELECT * FROM job_post";
if (!empty($_GET['experience'])) {
$sql = $sql." WHERE experience = '$_GET[experience]'";
}
if (!empty($_GET['qualification']) && !empty($_GET['experience'])) {
$sql = $sql." AND qualification = '$_GET[qualification]'";
}else if(!empty($_GET['qualification'])){
$sql = $sql." WHERE qualification = '$_GET[qualification]'";
}
$result=$conn->query($sql);
if ($result->num_rows>0) {
while($row=$result->fetch_assoc()){
$json[] = array(
0 => $row['jobtitle'],
1 => $row['description'],
2 => $row['minimumsalary'],
3 => $row['maximumsalary'],
4 => $row['experience'],
5 => $row['qualification'],
);
}
echo json_encode($json);
}
<section>
<div class="container">
<div class="row">
<div class="col-md-12">
<form id="myForm" class="form-inline">
<div class="form-group">
<label>Experience</label>
<select id="experience" class="form-control" name="exp">
<option disabled selected>Select Experience</option>
<?php
$sql = "SELECT DISTINCT(experience) FROM job_post WHERE experience is not null";
$result=$conn->query($sql);
if ($result->num_rows>0) {
while ($row = $result->fetch_assoc()) {
echo "<option value='".$row['experience']."'>".$row['experience']."</option>";
}
}
?>
</select>
</div>
<div class="form-group">
<label>Qualification</label>
<select id="qualification" class="form-control" name="qua">
<option disabled selected>Qualification</option>
<?php
$sql = "SELECT DISTINCT(qualification) FROM job_post WHERE qualification is not null";
$result=$conn->query($sql);
if ($result->num_rows>0) {
while ($row = $result->fetch_assoc()) {
echo "<option value='".$row['qualification']."'>".$row['qualification']."</option>";
}
}
?>
</select>
</div>
<button id="" class="btn btn-success">Search</button>
</form>
</div>
</div>
<div class="row" style="margin-top: 5%;">
<div class="table-responsive">
<table id="myTable" class="table" style="width: 1190px;">
<thead>
<th>Job Name</th>
<th>Job Description</th>
<th>Minimum Salary</th>
<th>Maximum Salary</th>
<th>Experience</th>
<th>Qualification</th>
</thead>
<tbody>
</tbody>
</table>
</div>
</div>
</div>
</section>
<script src="js/jquery-3.3.1.min.js"></script>
<script src="//cdn.datatables.net/1.10.19/js/jquery.dataTables.min.js"></script>
<script type="text/javascript">
$(function(){
var oTable = $('#myTable').DataTable({
"autoWidth" : false,
"ajax" : {
"url" : "refresh_job_search.php",
"dataSrc": "",
"data" : function(d){
d.experience = $("#experience").val();
d.qualification = $("#qualification").val();
}
}
});
$("#myForm").on("submit", function(e){
e.preventDefault();
oTable.ajax.reload(null, false);
});
});
</script>
<?php
session_start();
require_once("db.php");
$sql = "SELECT * FROM job_post";
if (!empty($_GET['experience'])) {
$sql = $sql." WHERE experience = '$_GET[experience]'";
}
if (!empty($_GET['qualification']) && !empty($_GET['experience'])) {
$sql = $sql." AND qualification = '$_GET[qualification]'";
}else if(!empty($_GET['qualification'])){
$sql = $sql." WHERE qualification = '$_GET[qualification]'";
}
$result=$conn->query($sql);
if ($result->num_rows>0) {
while($row=$result->fetch_assoc()){
$json[] = array(
0 => $row['jobtitle'],
1 => $row['description'],
2 => $row['minimumsalary'],
3 => $row['maximumsalary'],
4 => $row['experience'],
5 => $row['qualification'],
);
}
echo json_encode($json);
}
在初始化数据表之前，您需要先评估数据。数据表已经显示了数据库中的数据。问题是如果我搜索数据库中不存在的数据。它给出了一个警告。我希望更改该警告或显示一些数据不存在的消息更改PHP查询脚本以返回与表列匹配的空json数组（如果结果为0）。在这种情况下，DataTables错误将不会显示。您能告诉我怎么做吗？