我正在开发一个表单,它通过使用“jquery验证插件”和服务器端使用php验证客户端。 这是我的标记
HTML
<div id="recipe-form">
<div class="row">
<div class="col col-lg-9">
<form id="recipe-submit-form" class="form-horizontal" name="newad" method="post" enctype="multipart/form-data" action="<?php the_permalink(); ?>">
<div class="row control-group onoffclass">
<label for="recipetitle" class="col col-lg-2 control-label">Recipe Title:<sup>*</sup></label>
<div class="col col-lg-7 controls">
<input id="recipetitle" class="input-with-feedback" name="recipetitle" data-content="Required: Minimum 10 characters long" data-placement="top" data-toggle="popover" title="Recipe Title" placeholder="Recipe Title" type="text" />
<?php if($titleError != '') { ?><span class="nojserror"><?php echo $titleError;?></span><?php } ?>
</div>
</div> <!-- recipe title -->
<div class="row control-group onoffclass">
<label for="recipedesc" class="col col-lg-2 control-label">Recipe Desc:<sup>*</sup></label>
<div class="col col-lg-7 controls">
<textarea id="recipedesc" class="input-with-feedback" name="recipedesc" data-content="Required: A Brief recipe description" data-placement="top" data-toggle="popover" title="Recipe Description" placeholder="Recipe Short Description"></textarea>
<?php if($descError != '') { ?><span class="nojserror"><?php echo $descError;?></span><?php } ?>
</div>
</div> <!-- recipe desc -->
<div class="row">
<div id="submitform" class="col col-lg-10 col-offset-2">
<button name="Submit" type="submit" id="formsubmit" class="btn btn-default">Submit Recipe</button>
</div>
</div>
<input type="hidden" name="submitted" id="submitted" value="true" />
</form>
</div>
</div>
</div>
PHP
<?php
if(isset($_POST['submitted'])) {
//title
if(trim($_POST['recipetitle']) === '') {
$titleError = 'Please enter title for your recipe.';
$hasError = true;
} else if (strlen(trim($_POST['recipetitle']))<= 10) {
$titleError = 'Recipe Title is too short.';
$hasError = true;
} else {
$recipetitle = trim($_POST['recipetitle']);
}
//desc
if(trim($_POST['recipedesc']) === '') {
$descError = 'Please enter description for your recipe.';
$hasError = true;
} else if (strlen(trim($_POST['recipedesc']))<= 10) {
$descError = 'Recipe description is too short.';
$hasError = true;
} else {
$recipedesc = trim($_POST['recipedesc']);
}
}
?>
的jQuery
<script>
$(document).ready(function(){
var ruleSet1 = {
required: true,
minlength: 10
};
$('#recipe-submit-form').validate({
rules: {
recipetitle: ruleSet1,
recipedesc: ruleSet1,
}
});
});
</script>
每件事似乎工作正常,我面临的问题是,当我禁用Javascript时,php会显示验证错误。同时我启用Javascript并重新提交表单。现在jquery还将显示php错误旁边的验证错误。 请查看截图以了解该问题。
非常感谢您对此的任何帮助
此致
答案 0 :(得分:0)
根据js
或php
error
的具体情况,应用特定的错误类怎么办?
<span class="jsError">Some Error</span>
<span class="phpError">Some Error</span>
在css
中,您只需隐藏js
错误。
.jsError {
display:none;
}
在js
中,隐藏php errors
并在那里进行正常的jQuery验证,只会显示js errors
。
$(function() {
$('.phpError').hide();
//after doing your jsValidation show js error only
});
这样,如果禁用了js,则只会出现PHP
错误。