PHP從數組解析JSON

Question

我試圖從用json格式化的數組中取出一個鍵（狀態）。

陣列

Array ([0] => Running correlate.sh [1] => /var/www/html/blah/Correlation/classes [2] => Executing java command with JSON pps null 93993939 John Smith John Smith 1234 M 01-1-1990 [3] => last name is SMITH [4] => configFilePath is null [5] => feedName is pps [6] => [7] => select distinct(null), guid , scx_eid, emplid, national_ID, last_name, birthdate, first_name from table_a where guid is not null and (scx_eid = '12345678' or emplid = 'null' or (substr(national_ID,length(national_ID)-3,4) = '1234' and upper(last_name) = 'SMITH') or (substr(national_ID, length(national_ID)-3,4) = '1234' and birthdate = '90-null-01-1') or (birthdate = '90-null-01-1' and upper(last_name) = 'SMITH') or (upper(first_name) = 'JOHN' and upper(last_name) = 'ZSCHEILE')) [8] => first match {} [9] => [10] => select distinct(null), guid , eid, last4ssn, lastname, dob, firstname from table_c where guid is not null and (eid = '12345678' or (substr(last4ssn,length(last4ssn)-3,4) = '1234' and upper(lastname) = 'SMITH') or (substr(last4ssn, length(last4ssn)-3,4) = '1234' and dob = '90-null-01-1') or (dob = '90-null-01-1' and upper(lastname) = 'SMITH') or (upper(firstname) = 'JOHN' and upper(lastname) = 'SMITH')) [11] => second match {} [12] => [13] => select distinct(null), guid , eid, emplid from table_d where guid is not null and (eid = '12345678' or emplid = 'null') [14] => third match {} [15] => ldapMatch list is null [16] => returning firstMatch as {} [17] => [18] => Final Status:{Status=DISPUTED, GUID=[]} [19] => {"Status":"DISPUTED","GUID":[]}exiting script) 

我需要什麼

$status = 'DISPUTED'; 

更新

我能夠改變Java因此它輸出低於此。我怎麼能爆炸這個獲得兩個變量，$狀態和$ GUID？

{"Status":"DISPUTED","GUID":[]} 
{"Status":"CONFIRMED","GUID":[G010908712]} 
{"Status":"DISPUTED","GUID":[]} 
{"Status":"DISPUTED","GUID":[]} 
{"Status":"CONFIRMED","GUID":[G403057017]} 
{"Status":"CONFIRMED","GUID":[G104522939]} 
{"Status":"CONFIRMED","GUID":[G020669940]} 

Answer 1

如果字符串exiting script是真的在你的JSON會誤事的結束。否則，json_decode需要第二個參數來指定你想要一個數組，而不是一個對象：

$json_a = json_decode($array[19], true);