[英]Backup mysql Database take a specific table
您好,這是我用來備份mysql DATABASE的php代碼,它包含所有DATABASE和所有表,我想獲取數據庫中的特定表而不是所有表。 可能有人可以告訴我如何修改php代碼以獲取單個表而不是所有表嗎? 這是代碼:
<?php
/**
* Updated: Mohammad M. AlBanna
* Website: MBanna.info
*/
//MySQL server and database
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$dbname = 'novtech';
$tables = '*';
//Call the core function
backup_tables($dbhost, $dbuser, $dbpass, $dbname, $tables);
//Core function
function backup_tables($host, $user, $pass, $dbname, $tables = '*') {
$link = mysqli_connect($host,$user,$pass, $dbname);
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
exit;
}
mysqli_query($link, "SET NAMES 'utf8'");
//get all of the tables
if($tables == '*')
{
$tables = array();
$result = mysqli_query($link, 'SHOW TABLES');
while($row = mysqli_fetch_row($result))
{
$tables[] = $row[0];
}
}
else
{
$tables = is_array($tables) ? $tables : explode(',',$tables);
}
$return = '';
//cycle through
foreach($tables as $table)
{
$result = mysqli_query($link, 'SELECT * FROM '.$table);
$num_fields = mysqli_num_fields($result);
$num_rows = mysqli_num_rows($result);
$return.= 'DROP TABLE IF EXISTS '.$table.';';
$row2 = mysqli_fetch_row(mysqli_query($link, 'SHOW CREATE TABLE '.$table));
$return.= "\n\n".$row2[1].";\n\n";
$counter = 1;
//Over tables
for ($i = 0; $i < $num_fields; $i++)
{ //Over rows
while($row = mysqli_fetch_row($result))
{
if($counter == 1){
$return.= 'INSERT INTO '.$table.' VALUES(';
} else{
$return.= '(';
}
//Over fields
for($j=0; $j<$num_fields; $j++)
{
$row[$j] = addslashes($row[$j]);
$row[$j] = str_replace("\n","\\n",$row[$j]);
if (isset($row[$j])) { $return.= '"'.$row[$j].'"' ; } else { $return.= '""'; }
if ($j<($num_fields-1)) { $return.= ','; }
}
if($num_rows == $counter){
$return.= ");\n";
} else{
$return.= "),\n";
}
++$counter;
}
}
$return.="\n\n\n";
}
//save file , db-backup
//$fileName = 'novtechDB-'.time().'-'.(md5(implode(',',$tables))).'.sql';
$fileName = 'novtechDB'.'.sql';
$handle = fopen($fileName,'w+');
fwrite($handle,$return);
if(fclose($handle)){
echo "Done, the file name is: ".$fileName;
exit;
}
}
我將$ tables = array('myTable')放在表array [foreach($ tables as $ table)]的循環之前,我現在無法正常工作,但它給了我一個錯誤:注意:未定義的變量:return。 請有人可以幫助我找出問題所在嗎? 如何解決?
在表數組循環之前,將數組設置為要獲取的表。
$tables = ['tale1','table2','table3'];
foreach($tables as $table)
刪除注釋的代碼塊//get all of the tables
Bravooooooooooooooo! 我修復了它,謝謝#user3720435
步驟1:我刪除了注釋的代碼塊//獲取所有表
第2步:
我將$ tables = array('myTable')放在表array [foreach($ tables as $ table)]循環之前,我現在無法正常工作,但它給了我一個錯誤。 注意:未定義的變量:返回。
步驟3:我通過將return = null來解決它; 被注釋的代碼塊之前//表格
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