Que*_*ter 6 php forms firefox jquery
我使用以下脚本提交带有jQuery的表单,但Firefox已生成崩溃错误报告,请让我知道我做错了什么.
function teamValidation(){
var cat_id = $('#cat_id').val();
if(cat_id == "0"){
alert("Category is invalid...");
return false;
}else{
$('#addForm').submit();
return false;
}
}
<form action="<?php echo SITE_URL; ?>adminteams/saveDetail/"
name="addForm" id="addForm" onsubmit="return teamValidation();">
<input type="hidden" name="cat_id" id="cat_id" value="1"/>
</form>
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Firefox错误报告

它在其他浏览器中运行良好.
问题是您的表单多次提交.您可以使用以下任何一种方式实现目标:
HTML:
<form action="<?php echo SITE_URL; ?>adminteams/saveDetail/" name="addForm" id="addForm">
<input type="hidden" name="cat_id" id="cat_id" value="1"/>
</form>
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JAVASCRIPT:
$('#addForm').submit(function(){
var cat_id = $('#cat_id').val();
if(cat_id == "0"){
alert("Category is invalid...");
return false;
}else{
return true;
}
});
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另一种方式:
HTML:
<form action="<?php echo SITE_URL; ?>adminteams/saveDetail/" name="addForm" id="addForm" onsubmit="return teamValidation();">
<input type="hidden" name="cat_id" id="cat_id" value="1"/>
</form>
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JAVASCRIPT:
function teamValidation(){
var cat_id = $('#cat_id').val();
if(cat_id == "0"){
alert("Category is invalid...");
return false;
}else{
return true;
}
}
Run Code Online (Sandbox Code Playgroud)