在php中通过ajax显示数据
我想创建一个搜索函数。此页面已经有一些我从div中的DB中检索到的详细信息。我想在单击搜索按钮时隐藏所有检索到的详细信息并显示搜索到的详细信息。我使用了ajax。现在的问题是,当我单击“搜索”按钮时,以前检索到的数据会被隐藏,但不会显示为搜索结果在php中通过ajax显示数据,php,jquery,html,ajax,Php,Jquery,Html,Ajax,我想创建一个搜索函数。此页面已经有一些我从div中的DB中检索到的详细信息。我想在单击搜索按钮时隐藏所有检索到的详细信息并显示搜索到的详细信息。我使用了ajax。现在的问题是,当我单击“搜索”按钮时,以前检索到的数据会被隐藏,但不会显示为搜索结果 <form action="../PHP/searchrmvvac.php" method="post"> <div class="search hidden-xs hidden-sm"
<form action="../PHP/searchrmvvac.php" method="post">
<div class="search hidden-xs hidden-sm">
<input type="text" placeholder="Search" id="search" name="search">
</div>
<div>
<input type="button" value="Search" id="searchrmvcom" name="searchrmvcom">
<script>
$("#searchrmvcom").click(function () {
var comname=$('#search').val();
$.ajax({
type:"post",
url:"../PHP/searchrmvvac.php",
data:{comname:comname},
success:function (data3) {
$('#rmvcomdiv').hide();
$('#ela').html(data3)
}
});
});
</script>
</div>
</form>
$(“#searchrmvcom”)。单击(函数(){
var comname=$('#search').val();
$.ajax({
类型:“post”,
url:“../PHP/searchrmvvac.PHP”,
数据:{comname:comname},
成功:功能(数据3){
$('#rmvcomdiv').hide();
$('#ela').html(数据3)
}
});
});
searchrmvvac.php
<?php
session_start();
require('../PHP/dbconnection.php');
$output=$_POST['comname'];
$sql="select * from company where company_name='$output' and activation_code=1";
$res=mysqli_query($conn,$sql);
if(mysqli_num_rows($res)>0) {
echo '
while ($row = mysqli_fetch_assoc($res)) {
?>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.0/jquery.min.js"></script>
<table class="table table-striped table-bordered table-list">
<thead>
<tr>
<th>Action</th>
<th>ID</th>
<th>Registration number</th>
<th>Company Name</th>
<th>Email</th>
</tr>
</thead>
<tbody>
<tr>
<td align="center"><button type="submit" class="btn btn-default myButton" value="$row['/companyid/']" id="accept" name="accept">Remove</button></td>
<td>$row['/companyid/']</td>
<td>$row['/government_reg_no/']</td>
<td>$row['/company_name/']</td>
<td>$row['/email/']</td>
</tr>
</tbody>
</table>
<?php
}
';
}
?>[![enter image description here][1]][1]
[![在此处输入图像描述][1][1]
第27行是
<td align="center"><button type="submit" class="btn btn-default myButton" value="$row['/companyid/']" id="accept" name="accept">Remove</button></td>
删除
您的代码几乎没有问题,例如:
- 您的代码中没有包含/引用任何jQuery库,因此
首先不会起作用$(“#searchrmvcom”)。单击(…
- 在HTML表单中嵌入JavaScript/jQuery代码没有意义,请将它们分开
- searchrmvvac.php页面中没有JavaScript/jQuery代码,因此在这里包含jQuery库没有意义
- 您以错误的方式将PHP变量
嵌入到HTML表中。此外,您还以错误的方式执行了$row['…']
操作echo
<form action="../PHP/searchrmvvac.php" method="post">
<div class="search hidden-xs hidden-sm">
<input type="text" placeholder="Search" id="search" name="search">
</div>
<div>
<input type="button" value="Search" id="searchrmvcom" name="searchrmvcom">
</div>
</form>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.0/jquery.min.js"></script>
<script>
$("#searchrmvcom").click(function () {
var comname=$('#search').val();
$.ajax({
type:"post",
url:"../PHP/searchrmvvac.php",
data:{comname:comname},
success:function (data3) {
$('#rmvcomdiv').hide();
$('#ela').html(data3)
}
});
});
</script>
<?php
session_start();
require('../PHP/dbconnection.php');
$output=$_POST['comname'];
$sql="select * from company where company_name='$output' and activation_code=1";
$res=mysqli_query($conn,$sql);
if(mysqli_num_rows($res)>0) {
while ($row = mysqli_fetch_assoc($res)) {
?>
<table class="table table-striped table-bordered table-list">
<thead>
<tr>
<th>Action</th>
<th>ID</th>
<th>Registration number</th>
<th>Company Name</th>
<th>Email</th>
</tr>
</thead>
<tbody>
<tr>
<td align="center"><button type="submit" class="btn btn-default myButton" value="<?php echo $row['companyid']; ?>" id="accept" name="accept">Remove</button></td>
<td><?php echo $row['companyid']; ?></td>
<td><?php echo $row['government_reg_no']; ?></td>
<td><?php echo $row['company_name']; ?></td>
<td><?php echo $row['email']; ?></td>
</tr>
</tbody>
</table>
<?php
}
}
?>
$(“#searchrmvcom”)。单击(函数(){
var comname=$('#search').val();
$.ajax({
类型:“post”,
url:“../PHP/searchrmvvac.PHP”,
数据:{comname:comname},
成功:功能(数据3){
$('#rmvcomdiv').hide();
$('#ela').html(数据3)
}
});
});
随后,您的searchrmvvac.php页面将如下所示:
<form action="../PHP/searchrmvvac.php" method="post">
<div class="search hidden-xs hidden-sm">
<input type="text" placeholder="Search" id="search" name="search">
</div>
<div>
<input type="button" value="Search" id="searchrmvcom" name="searchrmvcom">
</div>
</form>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.0/jquery.min.js"></script>
<script>
$("#searchrmvcom").click(function () {
var comname=$('#search').val();
$.ajax({
type:"post",
url:"../PHP/searchrmvvac.php",
data:{comname:comname},
success:function (data3) {
$('#rmvcomdiv').hide();
$('#ela').html(data3)
}
});
});
</script>
<?php
session_start();
require('../PHP/dbconnection.php');
$output=$_POST['comname'];
$sql="select * from company where company_name='$output' and activation_code=1";
$res=mysqli_query($conn,$sql);
if(mysqli_num_rows($res)>0) {
while ($row = mysqli_fetch_assoc($res)) {
?>
<table class="table table-striped table-bordered table-list">
<thead>
<tr>
<th>Action</th>
<th>ID</th>
<th>Registration number</th>
<th>Company Name</th>
<th>Email</th>
</tr>
</thead>
<tbody>
<tr>
<td align="center"><button type="submit" class="btn btn-default myButton" value="<?php echo $row['companyid']; ?>" id="accept" name="accept">Remove</button></td>
<td><?php echo $row['companyid']; ?></td>
<td><?php echo $row['government_reg_no']; ?></td>
<td><?php echo $row['company_name']; ?></td>
<td><?php echo $row['email']; ?></td>
</tr>
</tbody>
</table>
<?php
}
}
?>
旁注:了解,因为现在您的查询容易受到SQL注入攻击。另请参见。您的代码很少有问题,例如:
- 您的代码中没有包含/引用任何jQuery库,因此
首先不会起作用$(“#searchrmvcom”)。单击(…
- 在HTML表单中嵌入JavaScript/jQuery代码没有意义,请将它们分开
- searchrmvvac.php页面中没有JavaScript/jQuery代码,因此在这里包含jQuery库没有意义
- 您以错误的方式将PHP变量
嵌入到HTML表中。此外,您还以错误的方式执行了$row['…']
操作echo
<form action="../PHP/searchrmvvac.php" method="post">
<div class="search hidden-xs hidden-sm">
<input type="text" placeholder="Search" id="search" name="search">
</div>
<div>
<input type="button" value="Search" id="searchrmvcom" name="searchrmvcom">
</div>
</form>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.0/jquery.min.js"></script>
<script>
$("#searchrmvcom").click(function () {
var comname=$('#search').val();
$.ajax({
type:"post",
url:"../PHP/searchrmvvac.php",
data:{comname:comname},
success:function (data3) {
$('#rmvcomdiv').hide();
$('#ela').html(data3)
}
});
});
</script>
<?php
session_start();
require('../PHP/dbconnection.php');
$output=$_POST['comname'];
$sql="select * from company where company_name='$output' and activation_code=1";
$res=mysqli_query($conn,$sql);
if(mysqli_num_rows($res)>0) {
while ($row = mysqli_fetch_assoc($res)) {
?>
<table class="table table-striped table-bordered table-list">
<thead>
<tr>
<th>Action</th>
<th>ID</th>
<th>Registration number</th>
<th>Company Name</th>
<th>Email</th>
</tr>
</thead>
<tbody>
<tr>
<td align="center"><button type="submit" class="btn btn-default myButton" value="<?php echo $row['companyid']; ?>" id="accept" name="accept">Remove</button></td>
<td><?php echo $row['companyid']; ?></td>
<td><?php echo $row['government_reg_no']; ?></td>
<td><?php echo $row['company_name']; ?></td>
<td><?php echo $row['email']; ?></td>
</tr>
</tbody>
</table>
<?php
}
}
?>
$(“#searchrmvcom”)。单击(函数(){
var comname=$('#search').val();
$.ajax({
类型:“post”,
url:“../PHP/searchrmvvac.PHP”,
数据:{comname:comname},
成功:功能(数据3){
$('#rmvcomdiv').hide();
$('#ela').html(数据3)
}
});
});
随后,您的searchrmvvac.php页面将如下所示:
<form action="../PHP/searchrmvvac.php" method="post">
<div class="search hidden-xs hidden-sm">
<input type="text" placeholder="Search" id="search" name="search">
</div>
<div>
<input type="button" value="Search" id="searchrmvcom" name="searchrmvcom">
</div>
</form>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.0/jquery.min.js"></script>
<script>
$("#searchrmvcom").click(function () {
var comname=$('#search').val();
$.ajax({
type:"post",
url:"../PHP/searchrmvvac.php",
data:{comname:comname},
success:function (data3) {
$('#rmvcomdiv').hide();
$('#ela').html(data3)
}
});
});
</script>
<?php
session_start();
require('../PHP/dbconnection.php');
$output=$_POST['comname'];
$sql="select * from company where company_name='$output' and activation_code=1";
$res=mysqli_query($conn,$sql);
if(mysqli_num_rows($res)>0) {
while ($row = mysqli_fetch_assoc($res)) {
?>
<table class="table table-striped table-bordered table-list">
<thead>
<tr>
<th>Action</th>
<th>ID</th>
<th>Registration number</th>
<th>Company Name</th>
<th>Email</th>
</tr>
</thead>
<tbody>
<tr>
<td align="center"><button type="submit" class="btn btn-default myButton" value="<?php echo $row['companyid']; ?>" id="accept" name="accept">Remove</button></td>
<td><?php echo $row['companyid']; ?></td>
<td><?php echo $row['government_reg_no']; ?></td>
<td><?php echo $row['company_name']; ?></td>
<td><?php echo $row['email']; ?></td>
</tr>
</tbody>
</table>
<?php
}
}
?>
旁注:了解,因为现在您的查询容易受到SQL注入攻击。另请参见。您的问题是什么?请清楚地解释您的问题,包括您的预期输出与当前输出等。是否存在代码缺失或键入错误?
echo'while($row=mysqli\u fetch\u assoc($res)){
->是否尝试回显循环?请使用错误报告(E\u ALL);ini\u集合('display\u errors',1)
在页面顶部,并让我们知道PHP告诉您的内容,如果出现错误…@RajdeepPaul没有错误,但单击“搜索”时不显示任何内容button@DanithKumarasinghe我在下面给出了答案。希望这能解决您的问题。您的问题是什么?请清楚地解释您的问题,包括您的预期输出与当前输出等代码丢失或键入错误?echo'而($row=mysqli\u fetch\u assoc($res)){
->是否尝试回显循环?请使用错误报告(E\u ALL);ini\u集('display\u errors',1)
在页面顶部,并让我们知道PHP告诉您的内容,如果出现错误…@RajdeepPaul没有错误,但单击“搜索”时不显示任何内容button@DanithKumarasinghe我在下面给出了一个答案。希望这能解决你的问题。