Question

我知道如何在json中使用url或使用普通参数传递数据，但这是不同的

见下图： -

http://www.tiikoni.com/tis/view/?id=23d9b98

我的api Url / geniedoc / api / doctor / Search

因为我们可以看到有一件事是在对象中而两个参数只是传递

你可以建议我打电话吗？

我的ajax电话

function getNewSlotContent(startIndex, totalRows) {
    var skipindex = 0;
    if (startIndex > 1) {
        skipindex = totalRows;
    }
    var totalrows = totalRows + 10;
    var searchByName = document.getElementById("searchByName").value;
    var selectedSpeciality = document.getElementById("searchByName").value;
    if (selectedSpeciality != null || selectedSpeciality != "") {
        selectedSpeciality = encodeURIComponent(selectedSpeciality);
    }
    $("#preloader").addClass("pageload");
    $("#preloader").show();

       var dataString = '{"firstName":"'+ searchByName + '","start_index":"' + skipindex + '","rows":"' + totalrows +'"}';

      console.log(dataString); 
    $.ajax({
         type: 'POST',
         url : '/geniedoc/api/doctor/search',
         data: dataString,
         contentType: 'application/json',
         dataType: 'json',
         headers: {'Content-Type':'application/json'}, 
        timeout: 100000,
        success: function(data) {
            var op = "";
            for (doctor in data.response.rows) {
                op += '<div class="post-sec">';
                op += '<a href=""><img src="/geniedoc/ajax/data/displayImage?fileName=' + data.response.rows[doctor].profilePicId + '" alt=""></a>';
                if (data.response.rows[doctor].prefix == null) {
                    data.response.rows[doctor].prefix = "";
                }
                op += '<a target="_blank" href="/doctor/' + data.response.rows[doctor].seo_name + '/' + data.response.rows[doctor].idKey + '" class="title"> ' + data.response.rows[doctor].prefix + ' ' + data.response.rows[doctor].firstame + ' ' + data.response.rows[doctor].last_name + '</a>';
                op += '<span class="date">' + data.response.rows[doctor].speciality_id + '</span>';
                op += '</div>';
                op += '<div class="clear"></div>';
            }
            $("#preloader").hide();
            $("#preloader").removeClass("pageload");
            $("#doctor-data").html(op);
        },
        error: function(e) {
            console.log("ERROR: ", e);
        },
        done: function(e) {
            console.log("DONE");
        }
    });

}

Answer 1

像这样修改你的dataString：

var dataString = {
"firstName":searchByName,
"start_index": skipindex,
"rows":  totalrows 
};

将对象和值传递给url时出现400错误请求

1 个答案: