我正在尝试触发由PHP变量触发的jquery .show()事件。
这是我的.php(从单独页面上的表单提出绘制):
$last_pic_displayed = trim($_POST["last_pic_displayed"]);
if ( strlen($last_pic_displayed) <= 0 )
{ $last_pic_displayed = trim($_GET["last_pic_displayed"]); }
这是我的.show()代码:
<script>
var lpic = "<?php echo $last_pic_displayed; ?>";
lpic = parseInt(lpic);
if (lpic == 0) {
$("#1").show();
}
</script>
我将此脚本上的变体用于多个Divs,目的是正确$ last_pic_display;值将触发正确的div显示,但到目前为止没有。
这是将数据发布到我的php的表格的示例:
<form id="buttons" action="process_form.php" method="post">
<label for="child_name"><img src="images/enter_name_text.png" alt="Enter Your Name" id="name_text"></label>
<input type="text" id="child_name" name="child_name" />
<section>
<input checked="checked" id="radio1" name="last_pic_displayed" type="radio" value="0" />
<label for="radio1"><img src="images/3_letter_text.png" alt="3 Letter Words" class="level_text">
</label>
</section>
<section>
<input checked="checked" id="radio2" name="last_pic_displayed" type="radio" value="6" />
<label for="radio1"><img src="images/4_letter_text.png" alt="4 Letter Words" class="level_text">
</label>
</section>
<section>
<input checked="checked" id="radio3" name="last_pic_displayed" type="radio" value="13" />
<label for="radio1"><img src="images/5_letter_text.png" alt="5 Letter Words" class="level_text">
</label>
</section>
<section id="submit_button">
<input type="image" name="submit" src="images/submit.png" />
</section>
</form>
这是我试图显示的代码部分的一个示例
<div class="word" id="1" data-value="1">
<img src="images/Bat.png" class="letter_text">
<form id="f1" action="results.php" method="post">
<input type="hidden" name="child_name" value="<?php $child_name; ?>" />
<input type="hidden" name="running_score" value="<?php $running_score; ?>" />
<input type="hidden" name="last_pic_displayed" value="1" />
<input type="hidden" name="correct_spelling" value="BAT" />
<input type="hidden" name="last_pic_flag" value="NO" />
<input type="hidden" name="level" value="3_letter" />
<input type="text" name="submitted_spelling" />
<input type="image" class="submit" src="images/submit.png" />
</form>
</div>
<script>
var lpic = "<?php echo $last_pic_displayed; ?>";
lpic = parseInt(lpic);
if (lpic == 0) {
$("#1").show();
}
alert(lpic);
</script>
我很新,所以我绝对感谢您的帮助。
编辑:用语法更正更新;问题当前持续存在。
edit2:用示例更新了。
edit3:用Parseint()更新和我要显示的Div的示例。
您需要parseInt()这样的" lpic"变量:
var lpic = "<?php echo $last_pic_displayed; ?>"; lpic = parseInt(lpic); if (lpic == 0) { $("#1").show(); }
use it:-
var lpic = "<?php echo $last_pic_displayed; ?>";
lpic = parseInt(lpic);
if (lpic == 0) {
$("#1").show();
}
使用此
<script>
var lpic = "<?php echo $last_pic_displayed; ?>"; // this must in ""
if (lpic == '0') { // must be == instead =
$("#1").show();
}
</script>