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[英]INSERT and UPDATE(+ increase values) between two different tables php mysql
[英]PHP/MySQL Update and Insert based on two tables
我已經討論了幾天,並且一直處於絆腳石。 我正在嘗試從table2中選擇一個唯一的ID,並將其與table1中的ID進行匹配。 如果ID匹配,請更新table1中的行並從table2中刪除記錄。 如果不匹配,則將記錄插入到table1中。
我已經到了可以更新記錄並插入新記錄的地步,但是我似乎無法在插入之前刪除匹配的記錄,因此創建了重復記錄。 我在更新后嘗試了刪除聯接查詢,但是由於新的logon_id前面有一個字符,因此它在表2中不再匹配。
我之前進行過更新聯接以進行更新,但是我進行的查詢過多,因此嘗試使其保持簡單。
有什么建議嗎? 在這個游戲中還是個新手。
$table2_query = "SELECT * FROM table2";
$table2_result = mysql_query($table2_query);
$table2_count = mysql_num_rows($table2_result);
if($table2_count == 0) {
if(mysql_error()) {
echo 'Error: '.mysql_error();
}
}
while($table2_row = mysql_fetch_array($table2_result, MYSQL_ASSOC)) {
$check_number_length = strlen($table2_row['unique_id']);
if($check_number_length < 7) {
if(substr($table2_row['unique_id'], 0, 2) < 35) {
$logon_id = 'n' . $table2_row['unique_id'];
}else {
$logon_id = 'v' . $table2_row['unique_id'];
}
}else {
$logon_id = $table2_row['unique_id'];
}
// Set variables for insert query for creation of a new user record
$first_name = $table2_row['firstname'];
$last_name = $table2_row['lastname'];
$email_address = $table2_row['email'];
$duplicates_query = "SELECT * FROM table1 WHERE '$logon_id' = logon_id";
$duplicates_result = mysql_query($duplicates_query);
$duplicates_row = mysql_fetch_array($duplicates_result);
$duplicates_count = mysql_num_rows($duplicates_result);
if($duplicates_count == 0) {
if(mysql_error()) {
echo 'Error: '.mysql_error();
}
}else {
$update_records_query = "UPDATE table1 SET first_name='$first_name', last_name='$last_name', email_address='$email_address'";
$update_records_result = mysql_query($update_records_query);
$update_records_count = mysql_affected_rows();
echo $update_records_count;
}
$create_records_query = "INSERT INTO table1 (first_name, last_name, email_address) VALUES ('$first_name', '$last_name', '$email_address')";
$create_records_result = mysql_query($create_records_query);
$create_records_count = mysql_affected_rows();
if($create_records_count == 0){
if(mysql_error()){
echo 'Error: '.mysql_error();
}
}
echo $create_records_count . ' record(s) created.';
}
$res = mysql_query ("SELECT count(table1.login_id) AS count FROM table2 LEFT JOIN table1 ON table2.login_id = table1.login_id WHERE table2.login_id = \"".$login_id."\";") or die ("Error joining tables");
/*joins both tables together and selects both id's from tables */
$row = mysql_fetch_assoc($res); // should only return one row so grab first
if ($row['count'] > 0) { // check if id exists in both tables (duplicates)
// updates rows from one table into another
mysql_query('UPDATE table1,table2 SET table1.username = table2.username, table1.password = table2.password, table1.email = table2.email WHERE table1.login_id = "'.$login_id.'" AND table2.login_id = "'.$login_id.'";') or die ("error updating table1");
//delete old row
mysql_query('DELETE FROM table2 WHERE login_id = "'.$login_id.'";') or die("error deleting from table2");
}else { // if id doesn't exist in table1
mysql_query ("INSERT INTO table1(username,password,email) SELECT username,password,email FROM table2 where login_id = '".$login_id."';") or die ("error inserting into table1");
}
刪除table1上login_id開頭的n
UPDATE table1 set login_id = replace(login_id,"n","");
新的更新查詢,如果在login_id開頭刪除n
mysql_query ('UPDATE table1,table2 SET table1.username = table2.username, table1.password = table2.password, table1.email = table2.email WHERE table1.login_id = table2.login_id AND table1.login_id = "'.$login_id.'";');
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