我需要在MySQL中加载100多个CSV文件,因此我编写了一个脚本来执行此操作。除此之外,我还有以下代码片段,它导入每个CSV文件:
private function importFile(){
if($this->connection->beginTransaction()){
$transactionFailed = false;
$importStatement = $this->connection->prepare("
LOAD DATA INFILE :file
REPLACE INTO TABLE :table;
");
foreach($this->fetchFileList() as $file){
$executeSuccesfull = $importStatement->execute(array(
':file' => $this->path . DIRECTORY_SEPARATOR . $file,
':table' => $file
));
if(!$executeSuccesfull){
$transactionFailed = true;
$this->fetchDBError($importStatement);
$this->connection->rollBack();
break;
}
}
if(!$transactionFailed){
$this->connection->commit();
}
}
else{
$this->fetchDBError();
}
}
$this->path
指向所有CSV文件所在的路径(我检查过)。每个$file
都与其导入数据的表具有完全相同的名称,没有文件扩展名(其中任何一个都没有.csv
)。我使用的是运行在Windows XP之上的PHP 5.3.9,MySQL 5.0.27。所有这些都在我的本地机器上运行。
问题是,我发现了以下错误:
[ERROR 42000] — 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
''the_first_file'' at line 2 (1064)
The query issued was:
LOAD DATA INFILE :file
REPLACE INTO TABLE :table;
此代码是由我班级的另一种方法生成的:
private function fetchDBError($statement = null){
$errorCode = null;
$errorInfo = null;
$queryString = '';
if($statement){
$errorInfo = $statement->errorInfo();
$errorCode = $statement->errorCode();
$queryString = "<p>The query issued was:</p><pre>{$statement->queryString}</pre>";
}
else{
$errorInfo = $this->connection->errorInfo();
$errorCode = $this->connection->errorCode();
}
$this->success = false;
$this->message = "<p>[ERROR $errorCode] {$errorInfor[0]} — {$errorInfo[2]} ({$errorInfo[1]})</p>$queryString";
答案 0 :(得分:0)
一个或多个文件包含aphostophe,请尝试使用:
$filepath = $this->path . DIRECTORY_SEPARATOR . str_replace("'", "''", $file);
然后在查询参数中使用$ filepath。
这是mysql加载数据函数的一个例子:
LOAD DATA INFILE '$filepath' INTO TABLE test
FIELDS TERMINATED BY ',';
还尝试指定FIELDS TERMINATED BY。
在stackoverflow帖子中,我找到了关于准备功能和LOAD DATA的this。