ajax传递json参数

var pros = [];
for(var i = 1; i <= 2; i++) {
	
	var obj = {};
	obj.id = i;
	obj.age = i*20;
	pros = pros.concat(obj);
}
pros = JSON.stringify(pros);
console.log(pros);	

  

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转载自www.cnblogs.com/lovekingly/p/11236678.html