2016-11-07 57 views
0

我有1個JSON,但我不能使用用於循環我不能使用環路(PHP)得到JSON數組值

$string = [{"lt":"1","lot":["1","1","1","1","1"]},{"lt":"2","lot":["0","0","0","0","0"]},{"lt":"3","lot":["0","0","0","0","0","0"]}] 
$json = json_decode($string,true); 
for($i = 0;$i<count($json);$i++){ 
    for($j = 0;$j<count($json->lot);$i++){ 
     if($json->lot==0){ 
      echo $j; 
     } 
    } 
} 

得到一個值i得到這個錯誤:試圖得到非對象

的屬性
+0

$ JSON是數組,使用'$ JSON [ '很多']',而不是'$ json-> lot' –

回答

0

編輯代碼:

$json = [{"lt":"1","lot":["1","1","1","1","1"]},{"lt":"2","lot":["0","0","0","0","0"]},{"lt":"3","lot":["0","0","0","0","0","0"]}] 
$json = json_decode($json); 
for($i = 0;$i<count($json);$i++){ 
    for($j = 0;$j<count($json->lot);$i++){ 
     if($json->lot==0){ 
      echo $j; 
     } 
    } 

改變JSON在使用數組PHP json_decode解碼()。 json_decode

+0

對不起IM忘記添加這對於s​​tackoverflow –

0

你需要在你的json上使用json_decode。另外,請記住,json必須作爲字符串存儲在PHP中。

<?php 
$json = ['{"lt":"1","lot":["1","1","1","1","1"]}','{"lt":"2","lot":["0","0","0","0","0"]}','{"lt":"3","lot":["0","0","0","0","0","0"]}']; 

foreach($json as $data) { 
    $current = json_decode($data); 

    foreach($current->lot as $lot) 
     echo $lot; 
} 
0

用於json_decode()功能解碼json

<?php 
$json = '[{"lt":"1","lot":["1","1","1","1","1"]},{"lt":"2","lot":["0","0","0","0","0"]},{"lt":"3","lot":["0","0","0","0","0","0"]}]'; 
$json = json_decode($json); 
foreach($json as $val){ 
    foreach($val->lot as $lot) { 
     if($lot==0){ 
      echo $lot; 
     } 
    } 
} 
?>