Question

我使用before submit钩子以便在提交之前运行一些javascript。因此，我使用以下代码：

jQuery(function() {
    jQuery('#setup-form').submit(function(e) {
        codeAddress();
        return true;
    });
});

函数codeAddress()用数据填充两个隐藏的输入字段。实际上填充了数据（进入值属性）。但是在服务器端没有数据到达。这有什么问题？

供参考：

function codeAddress() {
    var address = document.getElementById('address').value;
    if(address){
        geocoder.geocode( { 'address': address}, function(results, status) {
            if (status == google.maps.GeocoderStatus.OK) {
                var location = results[0].geometry.location;
                var location2 = '' + location;
                location2 = location2.slice(1, location2.length - 1).split(', ');
                jQuery('#geo-lat').attr('value', location2[0]);
                jQuery('#geo-long').attr('value', location2[1]);

                map.setCenter(location);
                var marker = new google.maps.Marker({
                    map: map,
                    position: location
                });
            } else {
                alert('Geocode was not successful for the following reason: ' + status);
            }
        });
    }
}

Answer 1

我认为问题是在隐藏字段中设置值之前提交表单。您可以做的是将您的功能代码放在提交块中，并在设置隐藏字段后返回true，这样您的隐藏字段将在提交之前设置。

试试这个

jQuery(function() {
jQuery('#setup-form').submit(function(e) {
    var address = document.getElementById('address').value;
    var isHiddenFieldSet=0;//0 indicates hidden field value is not set
    if(address){
        geocoder.geocode( { 'address': address}, function(results, status) {
            if (status == google.maps.GeocoderStatus.OK) {
                var location = results[0].geometry.location;
                var location2 = '' + location;
                location2 = location2.slice(1, location2.length - 1).split(', ');
                jQuery('#geo-lat').attr('value', location2[0]);
                jQuery('#geo-long').attr('value', location2[1]);

                map.setCenter(location);
                var marker = new google.maps.Marker({
                    map: map,
                    position: location
                });
                isHiddenFieldSet=1;//Set value to 1 to indicate hidden field value is set
            } else {
                alert('Geocode was not successful for the following reason: ' + status);
            }
        });
    }
    if(isHiddenFieldSet){//Submit the form only if hidden field value is set
          return true;
    } else {
         return false;
    }
    });
 });

Answer 2

首先调用您的codeAddress函数，然后从该函数内部调用您的提交后设置值。像这样

function codeAddress() {
    var address = document.getElementById('address').value;
    if(address){
                .
                .
                .
                jQuery('#geo-lat').attr('value', location2[0]);
                jQuery('#geo-long').attr('value', location2[1]);
                jQuery('#setup-form').submit();
             ...
}

Answer 3

由于Ravi Thakkar和mry的答案不起作用（我评论了原因）我建议我自己的解决方案，但这是基于你的想法：

function codeAddress(submit) {
    var address = document.getElementById('address').value;
    if(address){
        geocoder.geocode( { 'address': address}, function(results, status) {
            if (status == google.maps.GeocoderStatus.OK) {
                var location = results[0].geometry.location;
                var location2 = '' + location;
                location2 = location2.slice(1, location2.length - 1).split(', ');
                jQuery('#geo-lat').attr('value', location2[0]);
                jQuery('#geo-long').attr('value', location2[1]);

                map.setCenter(location);
                var marker = new google.maps.Marker({
                    map: map,
                    position: location
                });

                if(submit) jQuery('#setup-form').submit();
            } else {
                alert('Geocode was not successful for the following reason: ' + status);
            }
        });
    } else {
        if(submit) jQuery('#setup-form').submit();
    }
}

因此，将html中的提交按钮更改为：

是必要的

<button type="button" class="btn btn-success btn-lg" onclick="codeAddress(true)">Submit</button>

jQuery添加的帖子变量不会发送到服务器

3 个答案: