我有一个看起来像这样的数据库：
"trucks" : {
"3705ec54-8a2e-4eb1-8bb9-ab2243645ac1" : {
"email" : "[email protected]",
"name" : "Sandwich Truck",
"phone" : "123 - 456 - 1234",
"provider" : "password",
"selfDescription" : "We serve delicious sandwiches at a moderate price. Cards Accepted.",
"userType" : "truck",
"website" : "www.sandwiches.com"
},
"54fea8cd-2203-46bd-aaf8-9d823e85313d" : {
"email" : "[email protected]",
"name" : "Supa Pizza",
"phone" : "619 - 222 - 4444",
"provider" : "password",
"selfDescription" : "We serve incredible pizza at an incredibly unfair price.",
"userType" : "truck",
"website" : ""
},
"6c542367-507c-4d01-af2c-bf93a7efaef4" : {
"email" : "[email protected]",
"name" : "Pete's Fries",
"phone" : "11111111111111111111",
"profilePhoto" : "",
"provider" : "password",
"selfDescription" : "We make some of the world's most delicious fries.",
"userType" : "truck",
"website" : ""
},
"7c6c4395-aec1-443c-908d-62db517def5e" : {
"email" : "[email protected]",
"name" : "Mark's Chili",
"phone" : "1-800-CHIL-LLL",
"profilePhoto" : "",
"provider" : "password",
"selfDescription" : "We serve the most delicious chili, chili your mamma's mamma is scared to try.",
"userType" : "truck",
"website" : ""
}
}
我想实现一个搜索栏，将返回名称与搜索词匹配的卡车。例如，任何包含字符串“ fries”的卡车名称都将其ID提交给搜索结果ID数组。
到目前为止，这是我尝试过的方法，但没有骰子。当我在搜索栏中键入薯条并按下搜索按钮时，它不会打印“ 6c542367-507c-4d01-af2c-bf93a7efaef4”
let usersRef = Firebase(url: "https://•••••••••.firebaseIO.com/users")
usersRef.queryOrderedByChild("name").queryEqualToValue(searchBar.text).observeEventType(.ChildAdded, withBlock:{
snapshot in
print(snapshot.key)
})
我什至不知道我有多近，任何帮助将不胜感激。
谢谢！
最佳答案
您正在使用queryEqualToValue（Query.equalTo()）进行查询，只有存在完全匹配时，它才会返回结果。
详细说明：
如果在搜索栏中输入“炸薯条”，它将查找“炸薯条”的完全匹配项，该匹配项在我们的文档中不可用，因此未打印所需的值“ 6c542367-507c-4d01-af2c-bf93a7efaef4”。
相反，如果我们在搜索栏中输入值“ Pete's Fries”，则结果将为“ 6c542367-507c-4d01-af2c-bf93a7efaef4”
如果我们为搜索提供部分价值，那么我们正在尝试实现类似于SQL中的LIKE Query的搜索。请参考以下帖子以获取有关“如何在Firebase中执行LIKE查询”的更多信息
How to perform sql "LIKE" operation on firebase?