搜索在PHP

Question

JAVASCRIPT

function ShowContent(id) { 
if(id.length < 1) { return; } 
document.getElementById(id).style.display = "block"; 
} 

PHP标准

$test = mysql_query('SELECT userid,testid,testname FROM test_table WHERE userid = 9 ORDER BY name'); 
TestName : <select> 
      while($row=mysql_fetch_array($test)) 
      { 
       $testname = $row['testname']; 
       $testid = $row['testid']; 
       echo '<option value='.$testid.'>'.$testname.'</option> 
      } 
      </select> 
    echo '<div id="testid" style="display:none;"> 
     echo '<table> 
      <tr><th>USERID</th</tr> 
      <tr><th>TESTNAME</th></tr> 
      <tr><th>MARKS</th></tr> 
     publishtest = mysql_query('SELECT id,userid,publishtest,marks FROM publish_table WHERE userid= 9 GROUP BY id'); 
     while($row=mysql_fetch_array($publishtest)) 
     { 
     $userid = $row['userid']; 
     $testid = $row['id']; /*This is test id*/ 
     $testname = $row['publishtest']; /* This is test name*/ 
     $marks = $row['marks']; 
     echo '<tr><td>'.$userid.'</td> 
       <td>'.testid.'</td> 
       <td>'.testname.'</td> 
       <td>'.$marks.'</td> 
      </tr> 
     } 
     echo '</table>'; 
     echo '</div>'; 

，当我从选择标签测试我的需要，然后DIV应显示该testid的。我想无论是在JavaScript中，jQuery的

Answer 1

你可以试试这个：

的JavaScript

$('#showDivSelect').change(function (e) { 
    var divId = $(this).children('option:selected').val(); 
    $('#' + divId).css('display', 'block'); 
});​ 

HTML

<select id="showDivSelect"> 
<option id="o1">Div1</option> 
<option id="o2">Div2</option> 
</select> 
<div id="Div1"></div> 
<div id="Div2"></div>​ 

CSS

#Div1 { 
    width: 30px; 
    height: 30px; 
    background-color: red;   
    display: none; 
} 
#Div2 { 
    width: 30px; 
    height: 30px; 
    background-color: blue; 
    display: none;  
}​ 

这里是工作示例：http://jsfiddle.net/ePSDu/