Question

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循环来创建这样的表：

$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()); 

}

然后我使用另一个for循环将http://www.coinchoose.com/api.php的json api中的值插入到mysql表中，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

我希望有人能澄清我为什么会收到这个错误。任何建议，以更好地编写上述代码，因为我相信它可以以更漂亮/更好的方式完成，我们非常感激。代码运行，因为值被解析为mysql！但是，错误仍然出现

Answer 1

你的for循环中的条件错误。

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

或者，正如comments中所建议的那样：

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

Answer 2

在$results = json_decode($contents, true);插入$numcount = count($results);而不是for ($i=0; $i<=22; $i++){后，您应该插入如下变量：for ($i=0; $i<$numcount; $i++){

mysql插入错误json解码

2 个答案: