JS在执行隐藏/显示tr时单独打印第一行

时间:2018-07-14 10:51:45

标签: javascript php mysql

我有两个桌子。我想实现子行的显示/隐藏。我可以通过php获取相应的主行和子行。例如

ID      Field1   Field2

01        xxx      xxx

01        yyy      yyy

02        zzz      zzz

02        abc       abc

02        xyz      xyz

上表结构01 xxx 02 zzz zzz来自table1,其他来自表2。每个主行都有一个按钮。当我单击按钮时,将出现相应的子行。问题是当我单击按钮JS时仅返回第一行。我坚持下去。救救我

我的PHP代码:(抱歉,巨大的代码)

$connect = mysqli_connect('localhost', 'root', '', '');
$select1 =mysqli_query($connect,"SELECT * FROM ai");


while ($rownew=mysqli_fetch_array($select1)) 
{

?>
<input type='button' id="row_show_button<?php echo $rownew['idactionitems'];?>" value=<?php echo $rownew['idactionitems']?> onclick="ai_row_show_button('<?php echo $rownew['idactionitems'];?>');">
#some table datas here
</tr>
<?php 

 $new = mysqli_query($connect, "SELECT * FROM rowai WHERE idactionitemsFK =".$rownew['idactionitems']);


     while($newrow1=mysqli_fetch_array($new)) 
     {

?>

<tr style="display:" id="trhide<?php echo $newrow1['idactionitemsFK'];?>">
<td id="txt_name_edit1<?php echo $newrow1['idactionitemsFK'];?>"><?php echo $newrow1['idactionitemsFK'];?></td>
<td id="txt_name_edit2<?php echo $newrow1['idactionitemsFK'];?>"><?php echo $newrow1['row'];?></td>
</tr>
<?php
         }
}

?>

JS代码:

function ai_row_show_button(idactionitems) {
    var trhide = document.getElementById("trhide"+idactionitems);
    //alert (trhide);
       if (trhide.style.display == "none") {
           <?php
$newsc = mysqli_query($connect, "SELECT * FROM rowai WHERE idactionitemsFK = idactionitemsFK ");

    #
     while($newrow1=mysqli_fetch_array($newsc)) 
     {         
           ?>
        trhide.style.display = "block";
     <?php } ?>
         } else {
        trhide.style.display = "none";

    }
}

我要寻找的解决方案是单击第一个表生成的按钮时,将按钮的值传递给JS函数,并提供其中的所有<tr>与ID匹配的第二个表。目前,我仅收到第一行。请帮助我

0 个答案:

没有答案