英文:
Go: truthy condition not executing
问题
我正在为我的一个应用程序使用neo4j。
在运行查询后，如果找到了值，result.Next()会返回一个bool值。
var matches []int
fmt.Println(result.Next(), "<== result NEXT ???") // 👨💻 this prints true
if result.Next() {
// 👍 for some reason this block won't run!
fmt.Println("Assigning the values to the var")
matches = result.Record().Values()[0].([]int)
fmt.Println("Matches found", matches)
}
我真的很感谢帮助，已经卡在这个问题上几个小时了。
英文:
I am using neo4j in for one of my applications.
After running a query if the values are found, result.Next() returns a bool
var matches []int
fmt.Println(result.Next(), "<== result NEXT ???") // 👈 this prints true
if result.Next() {
// 👇 for some reason this block won't run!
fmt.Println("Assigning the values to the var")
matches = result.Record().Values()[0].([]int)
fmt.Println("Matches found", matches)
}
I would really appreciate the help, stuck on it for hours
答案1
得分: 5
调用result.Next()将进入下一行。如果调用两次，将跳过一行。result.Next()不是幂等的！如果只有一个结果，在调用result.Next()后，第二次调用将永远不会返回true。
如果您需要在多个位置检查result.Next()的结果，请将其存储在变量中：
var matches []int
hasNext := result.Next()
fmt.Println(hasNext, "<== result NEXT ???")
if hasNext {
fmt.Println("Assigning the values to the var")
matches = result.Record().Values()[0].([]int)
fmt.Println("Matches found", matches)
}
引用官方文档：消耗结果：
for result.Next() {
list = append(list, result.Record().Values[0].(string))
}
如您所见，通过简单调用result.Next()来迭代结果。
英文:
Calling result.Next() proceeds to the next row. If you call it twice, you skip a row. result.Next() is not idempotent! If you have only a single result, calling result.Next(), the second call will never return true.
If you need to examine the result of result.Next() at multiple places, store it in a variable:
var matches []int
hasNext := result.Next()
fmt.Println(hasNext, "<== result NEXT ???")
if hasNext {
fmt.Println("Assigning the values to the var")
matches = result.Record().Values()[0].([]int)
fmt.Println("Matches found", matches)
}
Quoting from the offical docs: Consuming results:
for result.Next() {
list = append(list, result.Record().Values[0].(string))
}
As you can see, results are iterated over by simply calling result.Next().
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