WordPress和SQL-如果列值不存在，则从另一个表更新和插入

Question

Stuck with some mysql for my wordpress site. All within 1 DB.

I am attempting to update wp_options table from wp_2_options (Multisite).

They both have option_name (unique_key) and option_value columns

It needs to update the option_value in the wp_options table if the option_name already exists in wp_2_options table

AND Insert the row from wp_2_options if the option_name doesnt exist.

So how do I UPDATE and `INSERT using the UNIQUE KEY value instead of the PRIMARY KEY

How do I do this?

Structure is the same for both of them

wp_options ( option_id bigint(20) unsigned NOT NULL AUTO_INCREMENT, option_name varchar(64) NOT NULL DEFAULT '', option_value longtext NOT NULL, autoload varchar(20) NOT NULL DEFAULT 'yes', PRIMARY KEY (option_id), UNIQUE KEY option_name (option_name) ) ENGINE=InnoDB AUTO_INCREMENT=1402 DEFAULT CHARSET=utf8;

Thanks in advance

---

UPDATE:

Ended up using this INSERT then UPDATE. This worked for me. Will just use this through PHP now.

-- Adding completely new rows from wp_2_options into wp_options INSERT IGNORE INTO wp_options SELECT * FROM wp_2_options WHERE option_name NOT IN (SELECT option_name FROM wp_options);

-- Updating the option_name field in wp_options, if that got changed in wp_2_options UPDATE wp_options JOIN wp_2_options ON wp_options.option_name = wp_2_options.option_name SET wp_options.option_value = wp_2_options.option_value;

Answer 1

There is nothing you need to do, the indexes will do the work for you.

Just do the INSERT and follow it with an UPDATE.

If it exists the INSERT will fail and the UPDATE will do the update.

If it does not exist the INSERT will put it there and the UPDATE will not affect any rows.

If I understand it correctly,
Something like this:

$sql = "SELECT `option_value`,`option_name` FROM `wp_options2` WHERE 1";
$result = mysqli_query($link, $sql);
while($row = mysqli_fetch_array($result, MYSQLI_NUM)){
  $sql = "INSERT INTO `wp_options`(`option_value`,`option_name`) VALUES( '$row[0]','$row[1]')";
  mysqli_query($link, $sql);
  $sql = "UPDATE `wp_options` SET `option_value` = '$row[0]' WHERE `option_name` = '$row[1]";
  mysqli_query($link, $sql);
}

If you need a warmer fuzzier feeling, use some error detection:

$sql = "SELECT `option_value`,`option_name` FROM `wp_options2` WHERE 1";
$result = mysqli_query($link, $sql);
while($row = mysqli_fetch_array($result, MYSQLI_NUM)){
  $sql = "INSERT INTO `wp_options`(`option_value`,`option_name`) VALUES( '$row[0]','$row[1]')";
  mysqli_query($link, $sql);
  $rows =  mysqli_affected_rows($link);
  $sql = "UPDATE `wp_options` SET `option_value` = '$row[0]' WHERE `option_name` = '$row[1]";
  $rows +=  mysqli_affected_rows($link);
  mysqli_query($link, $sql);
  if($rows == 0) {$errors[$row[0]] = $row[1];}
}
if(count($errors) > 0){
  echo '<p>Houston, we have a problem</p><pre>';
  var_export($errors);
  echo '</pre>';
}