Question

我必须按cscart_products文件csv更新表output.csv。

我创建了一个临时表（your_temp_table）并使用INNER JOIN命令更新了cscart_products表（product_code，amount）。

执行以下代码时，MySQL报告此错误：

用户更新失败：您的SQL语法出错;检查与您的MySQL服务器版本相对应的手册，以便在your_temp_table.amount = cs＆＃39;上使用接近＆＃39; UPDATE cscart_products INNER JOIN your_temp_table的正确语法。在第6行

有错误吗？ SQL语法错了吗？

<?php
$sql = "LOAD DATA INFILE 'output.csv'
       INTO TABLE your_temp_table
       FIELDS TERMINATED BY ','
    (product_code, amount);

UPDATE cscart_products
INNER JOIN your_temp_table on your_temp_table.amount = cscart_products.amount
SET cscart_products.product_code = your_temp_table.product_code;";

$con=mysqli_connect("localhost","xxx","xxxx","xxxx");
// Check connection
if (mysqli_connect_error()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
};

$result = mysqli_query($con, $sql);

if (mysqli_affected_rows($con) == 1) {
  $message = "The data was successfully added!";
} else {
  $message = "The user update failed: ";
  $message .= mysqli_error($con); 
};

echo $message;
mysqli_close($con);
?>

Answer 1

在mysqli中，它们应该作为两个查询执行

$sql = "LOAD DATA INFILE 'output.csv'
       INTO TABLE your_temp_table
       FIELDS TERMINATED BY ','
    (product_code, amount)";

$sql2 = "UPDATE cscart_products
INNER JOIN your_temp_table on your_temp_table.amount = cscart_products.amount
SET cscart_products.product_code = your_temp_table.product_code";


$result = mysqli_query($con, $sql);
$result = mysqli_query($con, $sql2);

或者，您必须使用multi_query

通过csv文件更新MySQL中的表

1 个答案: