Mysql插入错误json解码


mysql insert error json decode

我使用以下代码将json代码转换为数组并将值插入mysql。首先,我使用for循环创建如下表:

$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()); 
} 

然后我使用另一个for循环从http://www.coinchoose.com/api.php的json api插入值到mysql表,如下所示:

$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());  
}

我收到以下错误,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

我希望有人能澄清为什么我得到这个错误。任何建议,以更好的代码以上,因为我相信它可以在一个更漂亮/更好的方式完成非常感谢。代码确实工作,因为值被解析成mysql!然而,错误仍然出现

for循环条件错误。

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

或者,如注释所建议的:

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

$results = json_decode($contents, true);之后插入$numcount = count($results);,而不是for ($i=0; $i<=22; $i++){,您应该像这样插入变量:for ($i=0; $i<$numcount; $i++){