在php和mysql中自动完成文本框 [英] auto complete text box in php and mysql
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问题描述
< script src = assets / js / jquery-1.10.2.js > < / script >
< span class =code-keyword>< script src = assets / js / jquery-1.3.2.min.js > < / script >
< script 跨度> src = assets / js / jquery-ui-1.10.4.custom.min.js > < / script >
< script type = text / javascript >
jQuery( document )。ready( function (){
$(' #search')。autocomplete({
source:' search-job.php',
minLength: 2
});
});
< / script >
这里是我的HTML代码
< 输入 type = text id = 搜索 class = 表格 - 控制输入-lg col-xs-4 name = search > 跨度>
这里是我的search-job.php
<?php
包括' connectdb.php'< /跨度>;
if(isset($ _ REQUEST [' term']))
< span class =code-keyword> exit ();
$ rs = mysql_query( select * from jobs where title` like'% .ugfirst($ _ REQUEST [' term'])。 %'ORDER BY id ASC LIMIT 0,10)或 die(mysql_error( ));
$ data = array();
while($ row = mysql_fetch_assoc($ rs,MYSQL_ASSOC)){
$ data [] = array(
' label' => $ row [' title'],
' value' => $ row [' title']
);
}
echo json_encode($ data);
flush();
我在search-job.php中获得了正确的值。但它没有被html捕获但是当我运行代码时没有发生任何事情?为什么会这样?
这就是它在网络标签上的样子
图片
解决方案
(' #search')。autocomplete({
source:' search-job.php',
minLength: 2
});
});
< / script >
这里是我的HTML代码
< 输入 type = text id = 搜索 class = 表格 - 控制输入-lg col-xs-4 name = search > 跨度>
这里是我的search-job.php
<?php
包括' connectdb.php'< /跨度>;
if(isset(
_REQUEST [' term']))
exit ();
rs = mysql_query( 选择*来自`title`的工作,如'%'。ucfirst(
i`m trying to make a auto complete text box in php. here is my JavaScript code
<script src="assets/js/jquery-1.10.2.js"></script>
<script src="assets/js/jquery-1.3.2.min.js"></script>
<script src="assets/js/jquery-ui-1.10.4.custom.min.js"></script>
<script type="text/javascript">
jQuery(document).ready(function(){
$('#search').autocomplete({
source : 'search-job.php',
minLength:2
});
});
</script>
here is my html code
<input type="text" id="search" class="form-control input-lg col-xs-4" name="search">
here is my search-job.php
<?php
include 'connectdb.php';
if(isset($_REQUEST['term']))
exit ();
$rs= mysql_query("select * from jobs where `title` like '%" . ucfirst($_REQUEST['term']). "%' ORDER BY id ASC LIMIT 0,10") or die(mysql_error());
$data= array();
while($row= mysql_fetch_assoc($rs,MYSQL_ASSOC)){
$data[]=array(
'label'=>$row['title'],
'value'=>$row['title']
);
}
echo json_encode($data);
flush();
i`m getting correct values in search-job.php. but it is not catching by html but when i run code nothing happened? why is that?
this is how it's looks on network tab
image
解决方案
('#search').autocomplete({ source : 'search-job.php', minLength:2 }); }); </script>
here is my html code
<input type="text" id="search" class="form-control input-lg col-xs-4" name="search">
here is my search-job.php
<?php include 'connectdb.php'; if(isset(
_REQUEST['term'])) exit ();
rs= mysql_query("select * from jobs where `title` like '%" . ucfirst(
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