JSON到MySql使用PHP

Question

嗨，我需要你的帮助

我正在为类的项目工作，我正在寻找从JSON写入MySql数据库。

的JSON看起来像这样

{ 
"offset": 0, 
"results": [ 
    { 
     "aq1": 22, 
     "aq3": 27, 
     "aq2": 27, 
     "ateam/_source": "/nba/teams/page/NO/new-orleans-pelicans", 
     "aq4": 24, 
     "hqf/_source": "108", 
     "hq2": 27, 
     "hq1": 22, 
     "hq4": 36, 
     "hq3": 23, 
     "aqf/_source": "100", 
     "ateam/_text": "New Orleans", 
     "aq3/_source": "27", 
     "hq4/_source": "36", 
     "hteam": "http://www.cbssports.com/nba/teams/page/BOS/boston-celtics", 
     "aq2/_source": "27", 
     "hteam/_text": "Boston", 
     "aq1/_source": "22", 
     "hteam/_source": "/nba/teams/page/BOS/boston-celtics", 
     "ateam": "http://www.cbssports.com/nba/teams/page/NO/new-orleans-pelicans", 
     "aqf": 100, 
     "hq1/_source": "22", 
     "hq3/_source": "23", 
     "hq2/_source": "27", 
     "hqf": 108, 
     "aq4/_source": "24" 
    }, 

], 
"cookies": [], 
"connectorVersionGuid": "dgsadjgofsjkgkfsmghjfhf", 
"connectorGuid": "fbcjsdbgfjdjgkf", 
"pageUrl": "http://xsftbfbjjjsfaf.com/" 
} 

和我的PHP读取这样

<?php 
$server = "localhost"; 
$username = "lies"; 
$password = "madeup"; 
$database = "dgjsfdgjv"; 

mysql_connect($server,$username,$password) or die("Failed"); 
mysql_select_db($database) or die("Database Failed"); 

$url = "https://api.import.io/store/data/jjjjj-ec0f-4553-bda5-def61ca1756c/_query?input/webpage/url=http%3A%2F%2Fwww.jjjjj.com%2Fnba%2Fscoreboard%2F20150111&_user=f1508a78-05c5-4487-9219-51a990391329&_apikey=aLZu2wmRCvUlBwopi%2F6Kj%2F4wnscvZRh7DYQf80LE2e3hL22rtAp0nJ3lujy10cyx9JC9Ed73xb3zGp3aArhYDQ%3D%3D"; 
$string = file_get_contents($url); 
$arr = json_decode($string, true); 

foreach($arr as $item){ 
     $aq1 = $item['aq1']; 
     $aq3 = $item['aq3']; 
     $aq2 = $item['aq2']; 
     $ateam/_source = $item['aqeam/_source']; 
     $aq4 =$item['aq4']; 
     $hqf/_source =$item['aqf/_source']; 
     $hq2 =$item['aq2']; 
     $hq1 =$item['aq1']; 
     $hq4 =$item['aq4']; 
     $hq3 =$item['aq3']; 
     $aqf/_source =$item['aqf/_source']; 
     $ateam/_text =$item['aqeam/_text']; 
     $aq3/_source =$item['aq3/_source']; 
     $hq4/_source =$item['aq4/_source']; 
     $hteam =$item['aqeam']; 
     $aq2/_source =$item['aq2/_source']; 
     $hteam/_text =$item['aqeam/_text']; 
     $aq1/_source =$item['aq1/_source']; 
     $hteam/_source =$item['aqeam/_source']; 
     $ateam =$item['aqeam']; 
     $aqf =$item['aqf']; 
     $hq1/_source =$item['aq1/_source']; 
     $hq3/_source =$item['aq3/_source']; 
     $hq2/_source =$item['aq2/_source']; 
     $hqf =$item['aqf']; 
     $aq4/_source =$item['aq4/_source']; 



mysql_query("INSERT INTO createdb  (aq1,aq3,aq2,ateam/_source,aq4,hqf/_source,hq2,hq1,hq4,hq3,aqf/_source,ateam/_text,aq3/_source,hq4/_source,hteam,aq2/_source, hteam/_text,aq1/_source,hteam/_source,ateam,aqf,hq1/_source,hq3/_source,hq2/_source,hqf,aq4/_source) VALUES('$aq1','$aq3','$aq2','$ateam/_source','$aq4','$hqf/_source','$hq2','$hq1','$hq4','$hq3','$aqf/_source','$ateam/_text', '$aq3/_source','$hq4/_source','$hteam','$aq2/_source','$hteam/_text','$aq1/_source','$hteam/_source','$ateam','$aqf','$hq1/_source','$hq3/_source','$hq2/_source','$hqf','$aq4/_source')") or die ("Failed"); 
} 

我不触发任何错误消息出现，但在数据库中没有任何数据结束。我究竟做错了什么？它是正确的读取JSON字符串的PHP代码？

Answer 1

我明白了。

foreach ($phpArray['results'] as $key => $value) { 

// This is to make the data to be saved into the database 

foreach ($value as $k => $v) { 
     $aq1 = $value['aq1']; 
     $aq3 = $value['aq3']; 
     $aq2 = $value['aq2']; 
     $ateam_source = $value['ateam/_source']; 
     $aq4 =$value['aq4']; 
     $hqf_source =$value['hqf/_source']; 
     $hq2 =$value['aq2']; 
     $hq1 =$value['aq1']; 
     $hq4 =$value['aq4']; 
     $hq3 =$value['aq3']; 
     $aqf_source =$value['aqf/_source']; 
     $ateam_text =$value['ateam/_text']; 
     $aq3_source =$value['aq3/_source']; 
     $hq4_source =$value['hq4/_source']; 
     $hteam =$value['hteam']; 
     $aq2_source =$value['aq2/_source']; 
     $hteam_text =$value['hteam/_text']; 
     $aq1_source =$value['aq1/_source']; 
     $hteam_source =$value['hteam/_source']; 
     $ateam =$value['ateam']; 
     $aqf =$value['aqf']; 
     $hq1_source =$value['hq1/_source']; 
     $hq3_source =$value['hq3/_source']; 
     $hq2_source =$value['hq2/_source']; 
     $hqf =$value['hqf']; 
     $aq4_source =$value['hq4/_source']; 
} 

Answer 2

仔细看看你的问题，我意识到问题出在foreach() Loop之内。它应该是

foreach($arr as $item){ 
    echo $aq1 = $item[0]['aq1']; # note that [0] here 
    echo $aq3 = $item[0]['aq3']; 
    #...........Rest of Loop 
} 

而且也改名这样的变量$aqf/_source到$aqf_source，等等。