无法根据在搜索字段中输入的值显示表中的数据[关闭]
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【中文标题】无法根据在搜索字段中输入的值显示表中的数据[关闭]【英文标题】：Unable to show the data in table according to the value entered in serach field [closed] 【发布时间】：2020-06-09 13:03:05 【问题描述】： <input type="text" id="myInput" class="form-control" onkeyup="myFunction()" placeholder="Seach For Data..." title="Type in a name">
<table id="myTable" class="table table-bordered table-striped table-condensed cf">
<thead class="cf">
<tr id="headingTR" style="display:yes">
<th style="text-align:center">Application No.</th>
<th style="display:none;">ID</th>
<th style="width:15%;text-align:center" class="numeric">Name</th>
<th style="width:5%;text-align:center" class="numeric">Moblile No.</th>
<th style="width:45%;text-align:center" class="numeric">Detail</th>
<th style="width:10%;text-align:center" class="numeric">Tag</th>
<th style="width:5%;text-align:center" class="numeric">Status</th>
<th style="width:10%;text-align:center" >Download</th>
<th style="width:15%;text-align:center" class="numeric">Action</th>
</tr>
<tr id="noRecordTR" style="display:none">
<td>No Data</td>
</tr>
</thead>
<tbody id="data">
<?php
include("pagination/function1.php");
$page = (int) (!isset($_GET["page"]) ? 1 : $_GET["page"]);
$limit = 5; //if you want to dispaly 10 records per page then you have to change here
$startpoint = ($page * $limit) - $limit;
$sql_block = "SELECT * FROM citizen_request ORDER BY sno DESC LIMIT $startpoint, $limit";
$block_data = mysqli_query($con,$sql_block);
while($row = mysqli_fetch_assoc($block_data) )
$id = $row['sno'];
$name = $row['name'];
$mobile = $row['mobile'];
$description = $row['description'];
$remark = $row['remark'];
$uniqueid = $row['id'];
$status = $row['status'];
$file = $row['file'];
if ($file=="")
$file="blank.php";
?>
<tr>
<td data-title="Application No"><?=$uniqueid?></td>
<td data-title="ID" style="display:none;"><?=$id?></td>
<td class="numeric" data-title="Name"><?=$name?></td>
<td class="numeric" data-title="Mobile"><?=$mobile?></td>
<td class="numeric" data-title="Detail"><?=$description?></td>
<td class="numeric" data-title="Tag"><?=$remark?></td>
<td class="numeric" data-title="Status"> <span class="badge badge-primary"><?=$status?> </span></td>
<td class="numeric" data-title="Download"><a class=" btn btn-success btn-sm" href="upload/request/<?=$file?>" target=_blank> Download</a></td>
<td class="numeric" data-title="Action">
<a href="edit-request.php?sno=<?=$id?>" onclick="return confirmation()" ><button class="btn btn-primary btn-sm"><i class="fa fa-pencil"></i></button></a>
<a href="delete_request.php?sno=<?=$id?>" onclick="return confirmation()"> <button class="btn btn-danger btn-sm"><i class="fa fa-trash-o "></i></button></a>
</td>
</tr>
<?php
$sno=$sno+1;
?>
</tbody>
</table>
这是我的表格搜索字段。 这是我的脚本，用于获取用户搜索的数据。
$(document).ready(function()
$("#myInput").on("keyup", function()
var flag = false;
var value = $(this).val().toLowerCase();
$.ajax(
url: 'getRequestSearch.php',
type: 'post',
data: value:value,
dataType: 'json',
success:function(response)
var len = response.length;
if(len == 0)
else
var data = "";
var sno = 1;
for( var i = 0; i<len; i++)
var id = response[i]['id'];
var name = response[i]['name'];
var mobile = response[i]['mobile'];
var description = response[i]['description'];
var remark = response[i]['remark'];
var app_no = response[i]['app_no'];
var status = response[i]['status'];
var file = response[i]['file'];
if(file==""||file=="no")
file="blank";
else
if(id=="")
else
data += '<tr><td data-title="Application No.">' + app_no + '</td><td data-title="ID" style="display:none;">' + id + '</td> <td class="numeric" data-title="Name">' + name + '</td><td class="numeric" data-title="Mobile">'+ mobile + '</td><td class="numeric" data-title="Description">'+ description +'</td><td class="numeric" data-title="Remark">' +remark+ '</td><td class="numeric" data-title="Status"><span class="badge badge-primary">'+ status +'</span></td> <td class="numeric" data-title="Download"><a class=" btn btn-success btn-sm" href="upload/request/'+file+'" target=_blank>Download</a></td><td class="numeric" data-title="Operation"><a href="edit-request.php?sno='+id+'" onclick="return confirmation()" ><button class="btn btn-primary btn-sm"><i class="fa fa-pencil"></i></button></a><a href="delete_request.php?sno='+id+'" onclick="return confirmation()"> <button class="btn btn-danger btn-sm"><i class="fa fa-trash-o "></i></button></a></td></tr>';
sno++;
$("#download-excel").attr("href", "excel-request.php?status="+status_sel+"");
$("#download-excel").attr("target","_blank");
$("#download-zip").attr("href", "download-zip-request.php?status="+status_sel+"");
$("#download-zip").attr("target","_blank");
$("#data").html(data)
);
);
);
当我输入名称时，它会根据我在搜索字段中输入的内容获取数据作为响应。但表数据没有改变。 还有 m 使用 WHERE name LIKE %$value%;但是我如何搜索多个列或所有列。
【问题讨论】：
到目前为止，您做了什么来尝试调试问题？浏览器控制台是否显示任何错误？您是否尝试在调试器中单步执行您的 JS 代码？ “还有 m 使用 WHERE name LIKE %$value%;但是我如何搜索多个列或所有列。” - 通过为每列添加附加条件。 （如果这不是您想要的，那么也许可以查看全文搜索。） 你有onkeyup="myFunction()" 和 $("#myInput").on("keyup", function() 相同的元素为什么？
不，没有错误......而且我也得到了我正确的 json 数组。 @CBroe
如果我在搜索字段中输入“luf”，这是我的回复["id":"4","name":"Luffy","mobile":"8xxxx960","description":"Decription Is As Follows","remark":"forwarded to PM","app_no":"20200516141728","status":"\u0928\u093f\u0935\u0947\u0926\u0928 \u091c\u092e\u093e","file":"","id":"9","name":"Luffy","mobile":"81xxxx0","description":"Decription Is As Follows","remark":"forwarded to PM","app_no":"20200516141728","status":"\u0928\u093f\u0935\u0947\u0926\u0928 \u091c\u092e\u093e","file":""]
【参考方案1】：
$(document).ready(function()
$("#myInput").on("keyup", function()
var flag = false;
var value = $(this).val().toLowerCase();
$.ajax(
url: 'getRequestSearch.php',
type: 'post',
data: value:value,
dataType: 'json',
success:function(response)
var len = response.length;
if(len == 0)
else
var data = "";
var sno = 1;
for( var i = 0; i<len; i++)
var id = response[i]['id'];
var name = response[i]['name'];
var mobile = response[i]['mobile'];
var description = response[i]['description'];
var remark = response[i]['remark'];
var app_no = response[i]['app_no'];
var status = response[i]['status'];
var file = response[i]['file'];
if(file==""||file=="no")
file="blank";
else
if(id=="")
else
data += '<tr><td data-title="Application No.">' + app_no + '</td><td data-title="ID" style="display:none;">' + id + '</td> <td class="numeric" data-title="Name">' + name + '</td><td class="numeric" data-title="Mobile">'+ mobile + '</td><td class="numeric" data-title="Description">'+ description +'</td><td class="numeric" data-title="Remark">' +remark+ '</td><td class="numeric" data-title="Status"><span class="badge badge-primary">'+ status +'</span></td> <td class="numeric" data-title="Download"><a class=" btn btn-success btn-sm" href="upload/request/'+file+'" target=_blank>Download</a></td><td class="numeric" data-title="Operation"><a href="edit-request.php?sno='+id+'" onclick="return confirmation()" ><button class="btn btn-primary btn-sm"><i class="fa fa-pencil"></i></button></a><a href="delete_request.php?sno='+id+'" onclick="return confirmation()"> <button class="btn btn-danger btn-sm"><i class="fa fa-trash-o "></i></button></a></td></tr>';
sno++;
$("#download-excel").attr("href", "excel-request.php?status="+status"");
$("#download-excel").attr("target","_blank");
$("#download-zip").attr("href", "download-zip-request.php?status="+status"");
$("#download-zip").attr("target","_blank");
$("#data").html(data)
);
);
);
我只看到 status_sel 没有定义.. 现在它应该可以工作了
【讨论】：
应该如何工作？您在上面的代码中所做的哪些更改也添加了它？ 我提到了@Swati，我所做的只是将 status_sel 更改为 status，因为未定义 var status_sel 是的，它对我有用.. status_sel 是在其他功能中定义的，因为我猜它不起作用。以上是关于无法根据在搜索字段中输入的值显示表中的数据[关闭]的主要内容，如果未能解决你的问题，请参考以下文章
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