I found a nice example on here showing how to look through arrayObjects with a condition but I have a question.
As is stands its console.logging every time it the condition is false. Is it possible to only console.log once when its finished looping through everything.
var arrayObjects = [{
"building": "A",
"status": "good"
},
{
"building": "B",
"status": "horrible"
}
];
for (var i = 0; i < arrayObjects.length; i++) {
console.log(arrayObjects[i]);
for (key in arrayObjects[i]) {
if (key == "status" && arrayObjects[i][key] == "good") {
console.log(key + "->" + arrayObjects[i][key]);
} else {
console.log("nothing found");
}
}
}I'm assuming that you want it to print Nothing found when nothing's really found, not even a single thing..
Then, you can try this.
var arrayObjects = [{"building":"A", "status":"good"},
{"building":"B","status":"horrible"}];
var isFound = false;
for (var i=0; i< arrayObjects.length; i++) {
console.log(arrayObjects[i]);
for(key in arrayObjects[i]) {
if (key == "status" && arrayObjects[i][key] == "good") {
isFound = true
console.log(key + "->" + arrayObjects[i][key]);
}
}
}
if (isFound === false){
console.log("nothing found");
}
Simply use .length with if condition.
var arrayObjects = [{
"building": "A",
"status": "good"
},
{
"building": "B",
"status": "horrible"
}
];
for (var i = 0; i < arrayObjects.length; i++) {
console.log(arrayObjects[i]);
if( i === arrayObjects.length-1 ) {
console.log("nothing found");
}
}If you're willing to refactor you code, you can save on time complexity by using just one loop with Array.reduce()
var arrayObjects = [{
"building": "A",
"status": "good"
},
{
"building": "B",
"status": "horrible"
}
];
const foundKeys = arrayObjects.reduce((bool, key) => {
console.log(key)
if (key.status === "good") {
console.log("status ->", key.status);
bool = true
}
return bool
}, false)
if (!foundKeys) {
console.log("Nothing found")
}