我搜索过该网站并找到了一些示例，但并不是我的问题所特有的。
我正在用SQL创建一个frienship表，我想确保每一行都是唯一的。例如：如果用户x 是用户y ---＆gt;的朋友用户y 是x的朋友。 我试图让它无法进入（创建表格时）：
user1ID user2ID timeStamp
11 10 01/01/2010 00:00:00
10 11 01/01/2010 00:00:00
我写了这个：
CREATE TABLE FriendRelationship(
User1ID INTEGER NOT NULL,
User2ID INTEGER NOT NULL,
**CHECK (User1ID <> User2ID), /*User x cannot be a frien with himself*/
CHECK ((User1ID <> all User2ID) AND (User2ID <> all( User1ID))),
/*if user x friend with user y -> user y is friend with user x */**
FtimeStamp DATETIME NOT NULL,
FOREIGN KEY (User1ID) REFERENCES UserDF(UserID),
FOREIGN KEY (User2ID) REFERENCES UserDF(UserID),
PRIMARY KEY (User1ID, User2ID));
答案 0 :(得分：0)
我会采取不同的方法。由于哪个用户<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Demo</title>
<style>
#map {
height: 600px;
border: 1px solid #ccc;
}
.myCss{
text-align:center;
}
</style>
<link rel="stylesheet" href="https://unpkg.com/leaflet@1.0.3/dist/leaflet.css" />
<link rel="stylesheet" href="https://unpkg.com/leaflet.markercluster@1.0.3/dist/MarkerCluster.css" />
<link rel="stylesheet" href="https://unpkg.com/leaflet.markercluster@1.0.3/dist/MarkerCluster.Default.css" />
<link href='https://api.mapbox.com/mapbox.js/plugins/leaflet-fullscreen/v1.0.1/leaflet.fullscreen.css' rel='stylesheet' />
</head>
<body>
<div id="map"></div>
<script src="https://unpkg.com/leaflet@1.0.3/dist/leaflet.js"></script>
<script src="https://unpkg.com/leaflet.markercluster@1.0.3/dist/leaflet.markercluster-src.js"></script>
<script src='https://api.mapbox.com/mapbox.js/plugins/leaflet-fullscreen/v1.0.1/Leaflet.fullscreen.min.js'></script>
<script>
var addressPoints = [
[-31.953512, 115.857048, "<p class='myCss'><b>Morgan Stanley<br> Funder</b><p>"],
[-37.8210819833, 175.2213903167, "<p class='myCss'><b>Morgan Stanley<br> Funder</b><p>"],
[-34.206841, 142.136490, "<p class='myCss'><b>Morgan Stanley<br> Funder</b><p>"],
[-37.8211946833, 175.2213655333, "<p class='myCss'><b>Morgan Stanley<br> Funder</b><p>"]
];
</script>
<script type="text/javascript">
var tiles = L.tileLayer('http://{s}.tile.osm.org/{z}/{x}/{y}.png', {});
latlng = L.latLng(-25.2744, 133.7751);
var map = L.map('map', {center: latlng,
zoom: 4,
layers: [tiles],
fullscreenControl: {
pseudoFullscreen: false
}
});
var markers = L.markerClusterGroup();
for (var i = 0; i < addressPoints.length; i++) {
var a = addressPoints[i];
var title = a[2];
var marker = L.marker(new L.LatLng(a[0], a[1]), { title: title });
marker.bindPopup(title);
markers.addLayer(marker);
}
map.addLayer(markers);
</script>
</body>
</html>以及user1id无关紧要，我只能确定user2id必须始终是较小的数字（并且应用程序负责对这两个用户进行排序）插入时ID正确）。完成后，两者组合的主键将确保您没有重复的条目：
user1id