我很简单我的数据库中有2个表 1.发布 2.用户
Post表中的列是
User ID
Post Title
Post Content
User ID
然后是User表中的列
Post.User ID - 已连接到User Full Name
dataSource = new kendo.data.DataSource({
batch: true,
pageSize: 20,
transport: {
read: {
url: 'SaladEntry/GetSupport2/',
dataType: "json"
},
destroy: {
url: 'SaladEntry/DeleteSupportKendo2/',
type: "POST",
contentType: 'application/json'
},
create: {
url: 'SaladEntry/SaveSupportKendo2/',
type: "POST",
contentType: 'application/json',
complete: function (e) {
SupportGrid();
}
},
update: {
url: 'SaladEntry/EditSupportKendo2/',
type: "POST",
contentType: 'application/json',
complete: function (e) {
SupportGrid();
}
},
parameterMap: function (options, operation) {
if (operation == "read") {
return saladparamObj;
}
else {
options.models[0].CountryNo = $('#Country_No').val();
var SaladParamSerialized = JSON.stringify(options.models);
return SaladParamSerialized;
}
}
},
schema: {
model: {
id: "PK",
fields: {
CountryNo: { editable: true, nullable: true },
EffectiveDate: { type: "date" },
stringEffectiveDate: { type: "string" },
ScaleMin: { validation: { required: true } },
ScaleMax: { validation: { required: true } },
Currency: { type: "string" }
}
}
},
sort: {
field: "stringEffectiveDate",
dir: "desc",
compare: function (a, b, dir) {
return kendo.parseDate(a.EffectiveDate) - kendo.parseDate(b.EffectiveDate);
}
}
});
$("#grid").kendoGrid({
dataSource: dataSource,
pageable: true,
height: 300,
toolbar: ["create"],
columns: [
{
field: "stringEffectiveDate", title: "Date",
template: "#= kendo.toString(kendo.parseDate(EffectiveDate, 'yyyy-MM-dd'), 'yyyy-MM-dd') #", width: "100px",
sortable: {
allowUnsort: false,
compare: function (a, b, dir) {
return kendo.parseDate(a.EffectiveDate) - kendo.parseDate(b.EffectiveDate);
}
}
},
{ field: "EffectiveDate", title: "Effective Date", template: "#= kendo.toString(kendo.parseDate(EffectiveDate, 'yyyy-MM-dd'), 'yyyy-MM-dd') #", width: "100px" },
{ field: "ScaleMin", title: "BG1 Min", width: "100px" },
{ field: "ScaleMax", title: "BG7 Max", width: "100px" },
{ field: "Currency", title: "Currency", width: "100px" },
{ command: ["edit", "destroy"], title: "Commands", width: "160px" }
],
height: 300,
scrollable: true,
sortable: true,
filterable: true,
pageable: true,
resizable: true,
selectable: true,
editable: "popup",
cancel: function (e) {
nEdit();
},
edit: function(e) {
e.container.find(".k-edit-label:first").hide();
e.container.find(".k-edit-field:first").hide();
if (!e.model.isNew()) {
var dt = e.container.find("input[name=EffectiveDate]").data("kendoDatePicker");
dt.enable(false);
}
}
}).css("background-color", "#C7D6A7");
然后我现在要做的是选择Juan Cruz发布的帖子（User.User Full Name）。请帮助，我无法在任何地方找到这些问题。
PS抱歉我的英文不好。
编辑：我将在我的php / html项目中使用它作为我的搜索模块。所以我需要知道Juan Cruz发布的Juan Cruz用户ID的帖子。再次感谢
答案 0 :(得分：0)
您需要使用 JOIN 。这应该更新为使用您的列名。
SELECT p.id, p.title, p.content, u.full_name FROM User AS u JOIN Post AS p ON (u.id = p.user_id);
答案 1 :(得分：0)
希望这可能有所帮助：
SELECT
p.post_id, p.user_id, p.post_title, p.post_content,
u.user_fullname
// column name from two or more tables
// eg: table_name.column_name
// use p and u, because we already give temporary name for table post and user
FROM
post p // table 1: post refer as p (SQL aliases)
JOIN // combine 2 column from two or more tables
user u // table 2: user refer as u (SQL aliases)
ON
p.user_id = u.user_id // key for joining the table, the ID should be the same
或者您可以使用以下代码而不使用别名：
SELECT
post.post_id, post.user_id, post.post_title, post.post_content,
user.user_fullname
FROM
post
JOIN
user
ON
post.user_id = user.user_id
参考：SQL JOIN