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这是我的代码

var scrapURL = tgtURL+'&deptdate='+req.params.date+'&rtndate='+req.params.date+'&ddFrom='+req.params.from+'&ddTo='+req.params.to console.log(scrapURL)渗透 .get(scrapURL) .set({ 'oprname':["//span[@class='buscompanyname']"], 'oprimage':["//img[@class='buslogo']/@src "], 'dprtime':["//span[@class='bustime']"], 'pickup':["//span[@class='buspickup']"], 'dropoff':["/ /span[@class='busdropoff']"], 'coachtype':["//span[@class='bustype']"], 'price':["//span[@class='busprice'] "], })

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1 回答 1

0
This is my coding. Actually I want this coding connect to the database. But, i dont know how to create. Can you help me ? 

<script type="text/javascript">


        $(document).ready(function(){
            var tgtApi = 'http://localhost:3000/bus/';
            $('#tripForm').submit(function(e){
                // prevent form from reload
                e.preventDefault();
                // preparing url format
                tgtUrl = tgtApi+$('#from').val()+'&'+$('#to').val()+'&'+$('#when').val()
                // localhost:3000/bus/from&to&date url format
                console.log(tgtUrl); //F12 to view URL

                $.ajax(tgtUrl)
                .done(function(json){
                    console.log(json.data); //F12 to view result

                        for(var i=0; i<json.data[0].oprname.length; i++){
                        var list = '<tr><td><b>'+json.data[0].oprname[i]
                        +'<tr><td><li>Pickup:'+json.data[0].pickup[i]
                        +'<tr><td><li>Dropoff:'+json.data[0].dropoff[i]
                        +'<tr><td><li>Time:'+json.data[0].dprtime[i]
                        +'<tr><td><li>Coach Type:'+json.data[0].coachtype[i]
                        +'<tr><td><li>Price:'+json.data[0].price[i]
                        +'</tr></td></li><p>';
                        $('#triplist').append(list);
                    }
                });
            });
        });
xmlhttp.open("POST", "json_demo_db_post.php", true);
xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xmlhttp.send("x=" + dbParam);

</script>
于 2017-06-16T05:35:26.727 回答