我需要从多个API获取数据并在表中显示它们。
我想到的是:
var req = $.ajax({
url: "....linkhere",
dataType: 'jsonp'
});
var req = $.ajax({
url: "....linkhere1",
dataType: 'jsonp'
});
req.done(function(data) {
console.log(data);
var infoTable = $("<table />");
var arrayL = data.length;
var outputString = '';
for (var i = 0; i < arrayL; i++) {
var tableRow = $("<tr/>");
titleString = data[i].title;
var titleCell = $("<td />", {
text: titleString
});
detailString = data[i].description;
var detailCell = $("<td/>", {
text: detailString
});
tableRow.append(titleCell).append(detailCell);
infoTable.append(tableRow);
}
$("#display-resources").append(infoTable);
});
});
尽管如此,这样我只能从一个api获取数据。我怎样才能从多个数据库中获取它?
编辑:
$("#inputChoice").on("blur", function() {
let choice = $(this).val();
let req = $.ajax({
url: "...APIlink"+choice,
dataType: "jsonp"
});
req.done(function (data) {
console.log(data);
var infoTable = $("<table />");
let arrayL = data.length;
for (var i = 0; i < arrayL; i++) {
var tableRow = $("<tr/>");
titleString = data[i].title;
var titleCell = $("<td />", {
text: titleString
});
titleCell.addClass("title-row");
detailString = data[i].description;
var detailCell = $("<td/>", {
text: detailString
});
detailCell.addClass("details-row")
tableRow.append(titleCell).append(detailCell);
infoTable.append(tableRow);
}
$("#display-resources").append(infoTable);
});
});