php循环输出数组 json,php循环通过json数组(php loop through json array)
php循環(huán)通過(guò)json數(shù)組(php loop through json array)
我有一個(gè)像這樣的json字符串：
$fields_string = '
{"fields":
{"customers":[{"name":"john","id":"d1"},
{"name":"mike","id":"d2"},
{"name":"andrew","id":"d3"},
{"name":"peter","id":"d4"}]
}
}'
我該如何打印每個(gè)名字？ 稍后我將在html選項(xiàng)中使用它們，我知道如何做到這一點(diǎn)。 但是我無(wú)法把繩子拿出來(lái)。 這是我嘗試過(guò)的東西：
$obj = json_decode($fields_string);
$fields_detail = $obj-?{"fields"}->{"customers"};
在這一點(diǎn)上，我能夠通過(guò)echo json_encode($fields_detail)打印出客戶數(shù)組，但在此之前，我希望使用foreach來(lái)echo json_encode($fields_detail)名稱。 我試了好幾次，但沒(méi)辦法。 任何人都可以幫忙。
謝謝！
I have a json string like this:
$fields_string = '
{"fields":
{"customers":[{"name":"john","id":"d1"},
{"name":"mike","id":"d2"},
{"name":"andrew","id":"d3"},
{"name":"peter","id":"d4"}]
}
}'
How can I print each name? I will use them later in a html select options, I know how to do that. But I couldn't get the string out. Here are something I tried:
$obj = json_decode($fields_string);
$fields_detail = $obj-?{"fields"}->{"customers"};
at this point, I am able to print the customer array out by echo json_encode($fields_detail), but before that, I want to get the name break down using foreach. I tried several times, it didn't work. Can anyone help please.
Thanks!
原文：https://stackoverflow.com/questions/15000874
更新時(shí)間：2020-01-20 20:54
最滿意答案
客戶是一個(gè)對(duì)象數(shù)組，因此迭代每個(gè)對(duì)象并讀取屬性應(yīng)該有效。
foreach ($fields_detail as $customer) {
echo $customer->name;
}
Customers is an array of objects so iterating over each object and reading the property should work.
foreach ($fields_detail as $customer) {
echo $customer->name;
}
2013-02-21
相關(guān)問(wèn)答
如果需要關(guān)聯(lián)數(shù)組，請(qǐng)將第二個(gè)函數(shù)參數(shù)設(shè)置為true 如果您需要關(guān)聯(lián)數(shù)組，某些版本的php需要第2個(gè)參數(shù) $json = '[{"var1":"9","var2":"16","var3":"16"},{"var1":"8","var2":"15","var3":"15"}]';
$array = json_decode( $json, true );
Set the second function parameter to true if you require an associative a
...
還有一個(gè)級(jí)別高于您嘗試訪問(wèn)的級(jí)別。 嘗試這個(gè)： $tasks = $decoded['tasks'];
foreach($tasks as $task) {
echo $task["task"];
}
There is another level above what are you trying to access. Try this: $tasks = $decoded['tasks'];
foreach($tasks as $task) {
echo $task["task
...
json_decode()不返回?cái)?shù)組。 要做到這一點(diǎn)，你需要做json_decode($_REQUEST['names'], true) http://php.net/manual/en/function.json-decode.php json_decode() doesn't return an array. To get it to do so you'd need to do json_decode($_REQUEST['names'], true) http://php.net/manu
...
您是否有理由不使用json_encode來(lái)執(zhí)行此操作？ 您可以使用json_encode通過(guò)一個(gè)快速調(diào)用將普通的php數(shù)組轉(zhuǎn)換為JSON。 $arr = ("a"=>"one", "b"=>"two");
echo $json_encode($arr);
http://php.net/manual/en/function.json-encode.php Is there a reason you aren't using json_encode to do this? You can conv
...
你正在訪問(wèn)一個(gè)數(shù)組。 您使用[]執(zhí)行此操作，而不是使用{} 。 要獲得令牌數(shù)組，您需要$json2['reply1']['tokens'] 。 同樣，您使用[]訪問(wèn)此數(shù)組中的各個(gè)項(xiàng)目，而不是-> 。 使用$laa->token; 將再次失敗。 請(qǐng)嘗試以下方法 $tokens = $json2['reply1']['tokens'];
foreach ($tokens as $token) {
echo $token['token']
}
You're accessing an array.
...
在循環(huán)內(nèi)試試這個(gè)： if(isset($row['tableDet'])){// or !is_null() or is_numeric()
$tableDet = "occupied";
}else{
$tableDet = $row['tableDet']);//or = "free" or null
}
$contact= array(
"ID" => $row['tabledetailid'],
"Name" => $row['name'],
"tabl
...
一旦你在php中把一個(gè)關(guān)聯(lián)數(shù)組編碼成一個(gè)JSON對(duì)象，它不再是一個(gè)數(shù)組，它是一個(gè)帶有鍵和值的對(duì)象。 您可以使用for..in在javascript中迭代它們 for (var key in obj) {
console.log(obj[key]);
}
注意： for..in不是garuntee順序，如果你需要確保順序，你將不得不使用你的數(shù)組索引，而不是key=>value ，而是使用for循環(huán)(或while )。 Once you encode an associative array
...
客戶是一個(gè)對(duì)象數(shù)組，因此迭代每個(gè)對(duì)象并讀取屬性應(yīng)該有效。 foreach ($fields_detail as $customer) {
echo $customer->name;
}
Customers is an array of objects so iterating over each object and reading the property should work. foreach ($fields_detail as $customer) {
echo $custom
...
每次進(jìn)入循環(huán)時(shí)，都會(huì)覆蓋您的數(shù)組 嘗試： $json1[] = array(); $json2[] = array(); $this->db->sql = "SELECT ID AS ID, WebsiteName AS SelectText, WebsiteCode AS SelectValue FROM Banners_Websites ORDER BY WebsiteName";
$rs = $this->db->query($this->db->sq
...
您有許多具有相同數(shù)據(jù)的不必要變量。 你只需要做以下事情 <?php
header('Content-type: application/json');
$db = new PDO('sqlite:whad.db')or die('Could not open database');
$json = array(
"Daniel" => array(),
"Andi" => array(),
"Dave" => array(),
"Simon" => array()
);
for
...
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