我需要更改数组中字符串的所有实例,我可以这样做,但是当它们是键值对时,是不是更优雅的解决方案?
var person = [];
person.push({
name: 'John Smith',
description: 'John Smith is tall.',
details: 'John Smith has worked for us for 10 years.'
});
person.push({
name: 'Michael Smith',
description: 'Michael Smith is tall.',
details: 'Michael Smith has worked for us for 10 years.'
});
person.push({
name: 'Linda Smith',
description: 'Linda Smith is tall.',
details: 'Linda Smith has worked for us for 10 years.'
});
function replaceWordST() {
var toReplace = "Smith";
var replaceWith = "Jones";
for (i = 0; i < person.length; i++) {
person[i].name = person[i].name.replace(new RegExp(toReplace, 'g'), replaceWith);
person[i].description = person[i].description.replace(new RegExp(toReplace, 'g'), replaceWith);
// etcetc
}
}
答案 0 :(得分:2)
像对象一样: http://jsfiddle.net/mig1098/ekLmquap/
var replace = {
replaceWordST:function(person) {
var toReplace = /Smith/g;
var replaceWith = "Jones";
jsonData = person;
for(var obj in jsonData){
if(jsonData.hasOwnProperty(obj)){
for(var prop in jsonData[obj]){
if(jsonData[obj].hasOwnProperty(prop)){
jsonData[obj][prop] = replace.name(jsonData[obj][prop],toReplace,replaceWith);
}
}
}
}
return jsonData;
},
name:function(name,toReplace,replaceWith){
return name.replace(toReplace, replaceWith);
}
}
答案 1 :(得分:0)
更优雅的解决方案
您可以使用嵌套循环
来减少相似行的数量function replaceWordST() {
var toReplace = "Smith",
replaceWith = "Jones",
re = new RegExp(toReplace, 'g'), // cache this!!
keys = ['name', 'description', 'details' /*, etc*/],
i, j; // don't forget to var these!!
for (i = 0; i < theURLs.length; ++i) {
for (j = 0; j < keys.length; ++j) {
person[i][keys[j]] = person[i][keys[j]].replace(re, replaceWith);
}
}
}