所以基本上我在做,看起來像這樣(的var_dump)捲曲請求和響應越來越:PHP字符串JSON數組
string(595) "{"user_id":1,"currency":"eur","purchase_packs":{"1":{"amount":500,"allowed_payment_methods":["ideal","paypal","visa","mc"]},"3":{"amount":1000,"allowed_payment_methods":["mc","ideal","paypal","visa"]},"6":{"amount":2500,"allowed_payment_methods":["mc","ideal","paypal"]},"8":{"amount":5000,"allowed_payment_methods":["ideal"]},"9":{"amount":10000,"allowed_payment_methods":["ideal"]}},"payment_methods":{"ideal":{"name":"ideal","allow_recurring":false},"paypal":{"name":"paypal","allow_recurring":false},"visa":{"name":"visa","allow_recurring":false},"mc":{"name":"mc","allow_recurring":false}}}"
我想要的就是訪問它在這樣的JS文件:
success: function (data) {
alert(data.user_id);
}
但我不知道如何正確轉換(?)。
在那之後我的下一個步驟(問題)會,如果我能for循環每購買包做了,所以我可以爲他們的每一個
大概是這樣的創建按鈕:
var pack;
var packs = data.purchase_packs;
for (pack= 0; pack < packs.length; pack++) {
console.log(packs[pack]);
}
你的js ajax調用如果使用jquery可以有指定的響應類型,所以它會把你的php返回的json作爲本地對象存儲,然後你可以訪問。 – Dave
不需要在PHP中將其轉換爲任何內容......它已經是JSON字符串。只需要echo字符串以便消耗 – charlietfl
好吧,我已經嘗試過不轉換它,只是echo字符串,當我這樣做時:data.user_id它表示它是未定義的。 – user1410644