Question

我查看了我的代码并意识到脚本存在一些问题。现在我已经对它进行了分类，它将不会通过检查您是否拥有正确的资金。我认为这与我声明变量

的方式有关

<script type="text/javascript">
        var money = "<?php echo $ir['money']; ?>";
        var price = 5000;
        function openAccount() {
            var c = confirm("Are you sure you want to open a bank account for <?=money_formatter(price)?>?");
            if(c) {
                $('#main_text').html('Opening account'); 
                $('#main_text').append('Checking you have the correct funds'); 
                if(+money < +price) {
                    $('#main_text').append('<span style="color: red;">Incorrect funds, You need another <?=money_formatter((price)-$ir["money"])?>...</span>');         
                } else {
                    $.post('<?=file?>?open=true&print=1', function(data) {
                      if(data == 'not_enough_money') {
                        $('#main_text').append('Incorrect funds...');       
                      } else if(data == 'opened') {
                        $('#main_text').append('You have succesfully opened your brand new bank account!<a href="javascript:void(0);">&gt;View Account</a>');  
                      }
                    }); 
                }
            }
        }

Answer 1

你只是忘记了两个引号。

替换<a href="javascript:void(0); onClick=new_account();">
通过<a href="javascript:void(0);" onClick="new_account();">，一切都会奏效。

Javascript Void（0）不会调用函数

1 个答案: