<form id='myform'>
<input name='letra'value='a'/>
<input name='letra' value='b'/>
<input name='number1' value='1'/>
<input name='letra' value='d'/>
<input name='number2' value=''/>
<input name='letra' value='f'/>
</form>我的jquery是
var dataArray = $("#myform").serializeArray(),
len = dataArray.length;
for (i=0; i<len; i++) {
if(dataArray[i].name == 'number1' || dataArray[i].name == 'number2' ) {
if($.isNumeric(dataArray[i].value)){
alert('campo '+dataArray[i].name +' is number');
}else
alert('campo '+dataArray[i].name +' not number');
}else
alert('campo '+dataArray[i].name +' si tiene');
}控制台告诉我:
Uncaught:$.isNumeric不是函数
我能帮忙吗?是用于优化我的源代码c:,还是我不知道如何使用$.isNumeric??
发布于 2015-12-05 02:01:42
还有其他代码可以引起麻烦吗?美元冲突了吗?试试jQuery.isNumeric?
你发布的代码对我来说很好
var dataArray = $("#myform").serializeArray(),
len = dataArray.length;
for (i=0; i<len; i++) {
if(dataArray[i].name == 'number1' || dataArray[i].name == 'number2' ) {
if($.isNumeric(dataArray[i].value)){
alert('campo '+dataArray[i].name +' is number');
}else
alert('campo '+dataArray[i].name +' not number');
}else
alert('campo '+dataArray[i].name +' si tiene');
}<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<form id='myform'>
<input name='letra'value='a'/>
<input name='letra' value='b'/>
<input name='number1' value='1'/>
<input name='letra' value='d'/>
<input name='number2' value=''/>
<input name='letra' value='f'/>
</form>
https://stackoverflow.com/questions/34100195
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