var eachMapping = [{"fname":"John"},{"fname":"John"}];
var count = 0;
for(var i = 0; i < eachMapping.length; i++){
for(var prop in eachMapping[i]){
if(eachMapping[i][prop] = "fname"){
count+=1
}
}
}
console.log(count);
最佳答案
首先:使用单个“ =”更改eachMapping [i] [prop]的值,而不是检查相似性。尝试将其更改为“ ==“
第二:U不需要双循环。尝试:
for(var i = 0; i < eachMapping.length; i++){
if(eachMapping[i]["fname"]){
count+=1
}
}
关于javascript - 我在Javascript中有一个对象数组。如何遍历它以查找'fname'键的出现次数,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/42600975/