onchange选择做不同的mysql查询 [英] onchange select do different mysql query

问题描述

正如标题所说，我需要onchange帮助。我有选择标记，当我从选择列表中选择某些东西时，我需要执行不同的mysql查询。例子：

 < select name =list> 
< option>汽车< / option> 
< option> busses< / option> 
< option>卡车< / option> 
< / select> 
  

然后当我选择 cars 这个

  $ query =select * from table where type ='cars'; 
  

如果我选择卡车，它会执行

  $ query =select * from table where type ='trucksters'; 
  

等等...

然后我需要显示结果在列表下的div
示例

 < select name =list> 
< option>汽车< / option> 
< option> busses< / option> 
< option>卡车< / option> 
< / select> 
< div>这是我需要显示查询结果< / div> 
  

请帮忙!!!

解决方案

如果您想通过ajax更改结果：

HTML：

 < select id =listname =list> 
 
< option>汽车< / option> 
 
< option>巴士< / option> 
 
< option>卡车< / option> 
< / select> 
  

在JS文件中：

  $（document）.ready（function（）{
 $（'＃list'）。change（function（）{
 var selected = $（this）.val ）; 
 $ .get（change_query.php？selected =+ selected，function（data）{
 $（'。result'）。html（data）; 
 
}）; 
}）; 
}）; 
  

，您应该创建change_query.php文件并在其中查询和代码并返回结果

  $ type = $ _POST [selected]; 
 
 $ query =select * from table where type ='。$ type。'; 
  

print result here;;

• 告诉我你是否需要任何帮助，我只是在Jquery中引导你而不是所有代码

As title says I need help with onchange. I have select tag,and I need to do different mysql query when I choose something from select list. Example:

<select name="list">
<option>cars</option>
<option>busses</option>
<option>trucks</option>
</select>

and then when i select cars it do this

$query="select * from table where type='cars'";

and if I choose trucks it do

$query="select * from table where type='trucks'";

and so on...
then I need to display the result in div under the list example

<select name="list">
<option>cars</option>
<option>busses</option>
<option>trucks</option>
</select>
<div> this is where I need to display results from query</div> 

Please help!!!

解决方案

If you want to change the result by ajax :

HTML :

 <select id="list" name="list">

    <option>cars</option>

    <option>busses</option>

    <option>trucks</option>
    </select>

in JS File:

$(document).ready(function() {
  $('#list').change(function() {
var selected=$(this).val();
  $.get("change_query.php?selected="+selected, function(data){
      $('.result').html(data);

    });
    });
});

and you should create change_query.php file and your query and code in it and return the result in it

$type = $_POST["selected"];

$query="select * from table where type='".$type."'";

print result here .... ;

• Tell me if you need any help ,I just guided you in Jquery not all code

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