mysql插入錯誤json解碼

Question

我正在使用以下代碼將json代碼轉換爲數組並將值插入到mysql中。首先，我用一個for循環來創建這樣的表：

$url='http://www.coinchoose.com/api.php'; 
$contents = file_get_contents($url); 
$contents = utf8_encode($contents); 
$results = json_decode($contents, true); 

for ($i=0; $i<=22; $i++){ 

mysql_query("CREATE TABLE $symbol(
id INT NOT NULL AUTO_INCREMENT, PRIMARY KEY(id), timestamp BIGINT, symbol VARCHAR(3), name VARCHAR(20), algo VARCHAR(20), currentBlocks VARCHAR(20), difficulty DECIMAL (18,9), reward DECIMAL (18,9), price DECIMAL (18,9), exchange VARCHAR(20), ratio DECIMAL (8,4))") 
or die(mysql_error()); 

} 

然後我用另一個for循環插入來自JSON API值在http://www.coinchoose.com/api.php到MySQL表，就像這樣：

$url='http://www.coinchoose.com/api.php'; 
$contents = file_get_contents($url); 
$contents = utf8_encode($contents); 
$results = json_decode($contents, true); 
print_r($results); 

$time=time(); 

for ($i=0; $i<=22; $i++){ 
$symbol=strtolower($results[$i]['symbol']); 
$name=$results[$i]['name']; 
$algo=$results[$i]['algo']; 
$currentBlocks=$results[$i]['currentBlocks']; 
$difficulty=$results[$i]['difficulty']; 
$reward=$results[$i]['reward']; 
$price=$results[$i]['price']; 
$exchange=$results[$i]['exchange']; 
$ratio=$results[$i]['ratio']; 

mysql_query("INSERT INTO $symbol VALUES (id, $time, '$symbol', '$name', '$algo', '$currentBlocks', $difficulty, '$reward', $price, '$exchange', $ratio)") or die(mysql_error()); 

} 

我收到以下錯誤，我不明白：

(!) Notice: Undefined offset: 21 in C:\wamp\www\api.php on line 45 
Call Stack 
# Time Memory Function Location 
1 0.0004 706424 {main}() ..\api.php:0 

(!) Notice: Undefined offset: 21 in C:\wamp\www\api.php on line 46 
Call Stack 
# Time Memory Function Location 
1 0.0004 706424 {main}() ..\api.php:0 

(!) Notice: Undefined offset: 21 in C:\wamp\www\api.php on line 47 
Call Stack 
# Time Memory Function Location 
1 0.0004 706424 {main}() ..\api.php:0 

(!) Notice: Undefined offset: 21 in C:\wamp\www\api.php on line 48 
Call Stack 
# Time Memory Function Location 
1 0.0004 706424 {main}() ..\api.php:0 

(!) Notice: Undefined offset: 21 in C:\wamp\www\api.php on line 49 
Call Stack 
# Time Memory Function Location 
1 0.0004 706424 {main}() ..\api.php:0 

(!) Notice: Undefined offset: 21 in C:\wamp\www\api.php on line 50 
Call Stack 
# Time Memory Function Location 
1 0.0004 706424 {main}() ..\api.php:0 

(!) Notice: Undefined offset: 21 in C:\wamp\www\api.php on line 51 
Call Stack 
# Time Memory Function Location 
1 0.0004 706424 {main}() ..\api.php:0 

(!) Notice: Undefined offset: 21 in C:\wamp\www\api.php on line 52 
Call Stack 
# Time Memory Function Location 
1 0.0004 706424 {main}() ..\api.php:0 

(!) Notice: Undefined offset: 21 in C:\wamp\www\api.php on line 53 
Call Stack 
# Time Memory Function Location 
1 0.0004 706424 {main}() ..\api.php:0 


You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'VALUES (id, 1369998276, '', '', '', '', , '', , '',)' at line 1 

我希望有人能解釋，爲什麼我得到這個錯誤。任何建議，以更好地編碼上述，因爲我相信它可以做一個更漂亮/更好的方式非常讚賞。該代碼確實工作，因爲值被分析到MySQL！但是，錯誤仍然出現

Answer 1

您的for循環中有錯誤的條件。

for ($i = 0; $i < 22; $i++) { // Notice the `<` and not `<=` 

---

或者，如建議在comments：

for ($i = 0; $i < count($result); $i++) { 

Answer 2

後$results = json_decode($contents, true);插入$numcount = count($results);和代替for ($i=0; $i<=22; $i++){你應該插入這樣的變量：for ($i=0; $i<$numcount; $i++){