我创建了一个代码,通过点击另一个RADIO点击RADIO和其他信息来显示信息。
正常工作,但是当我更新页面时,即使是CHECKED,也没有显示任何信息。然后我必须再次点击已经显示CHECKED信息的RADIO。
任何人都知道如何解决?
请遵循以下代码:
<script>
// COLOR BG OR IMG BG
$(document).ready(function(){
$("input[name$='userBgOrImg']").click(function() {
var test = $(this).val();
$("div.desc").hide();
$("#"+test).show();
});
});
</script>
<style type="text/css">
.desc { display: none; }
</style>
<div><label><input type="radio" name="userBgOrImg" value="1" <?php $id = $res_select["id"]; if ( $userBgOrImg == 1 ) { echo "checked=\"checked\"";} else { echo "";}?>>Cor de Fundo</label></div>
<div><label><input type="radio" name="userBgOrImg" value="2" <?php $id = $res_select["id"]; if ( $userBgOrImg == 2 ) { echo "checked=\"checked\"";} else { echo "";}?>>Imagem de Fundo</label></div>
<div> <br /> <br /> </div>
<div id="1" class="desc">
<label for="userBgBody">Cor do fundo do Site</label>
<input type="text" class="form-control bscp span12" value="<?php $id = $res_select["id"]; echo $userBgBody; ?>" id="userBgBody" name="userBgBody">
<hr />
</div>
<div id="2" class="desc">
<label for="userImgBody">Imagem de fundo do site</label>
<input type="text" class="span12" name="userImgBody" id="userImgBody" placeholder="Imagem do fundo do site" value="<?php $id = $res_select["id"]; echo $userImgBody; ?>" />
<hr />
<label for="userImgBodyRepeat">Repetir imagem de fundo</label>
<span> <input type="radio" name="userImgBodyRepeat" value="repeat"> Sim</span> <span> <input type="radio" name="userImgBodyRepeat" value="no-repeat" selected="selected"> Não
<hr />
</div>
提前感谢您的帮助。
答案 0 :(得分:0)
试试这个,
$(document).ready(function(){
var did=$("input[name$='userBgOrImg']:checked").val();
if(did)
{
$("#"+did).show();
}
// your remaining code
});// document.ready function close
答案 1 :(得分:0)
你需要尝试
$(document).ready(function(){
$("input[name$='userBgOrImg']").change(function() {
var test = $(this).val();
$("div.desc").hide();
if(this.checked) {
$("#"+test).show();
}
}).filter(':checked').change();
});
答案 2 :(得分:0)
Tr this ..
$(document).ready(function () {
$("input[name$='userBgOrImg']").click(function () {
var value = $(this).val();
if(value)
{
$("#"+value).show();
}
});