Display all data of all tables

Question

I want to display all the data in my database without an entry selectfor each table - how can I do this?

I do not want to do this:

select * from Customer
select * from Employee
select .............

I am using TSQL with an MSSQL server.

Answer 1

DECLARE @sqlText VARCHAR(MAX)
SET @sqlText = ''
SELECT @sqlText = @sqlText + ' SELECT * FROM ' + QUOTENAME(name) + CHAR(13) FROM sys.tables
EXEC(@sqlText)

Answer 2

For mysql:

• Run SELECT information_schema.TABLES.TABLE_NAME FROM information_schema.TABLES where table_schema = 'db_name'

• Create a loop that will run the select query for each table retrieved from the first query.

Answer 3

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Answer 4

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, MySql phpMyAdmin - , , SQL .

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<?php
listAll("table_name");
function listAll($db) {
  mysql_connect("localhost","root","");
  mysql_select_db($db);
  $tables = mysql_query("SHOW TABLES FROM $db");
  while (list($tableName)=mysql_fetch_array($tables)) {
    $result = mysql_query("DESCRIBE $tableName");
    $rows = array();
    while (list($row)=mysql_fetch_array($result)) {
      $rows[] = $row;
    }
    $count = count($rows);
    if ($count>0) {
      echo '<p><strong>',htmlentities($tableName),'</strong><br /><table border="1"><tr>';
      foreach ($rows as &$value) {
        echo '<td><strong>',htmlentities($value),'</strong></td>';
      }
      echo '</tr>';
      $result = mysql_query("SELECT * FROM $tableName");
      while ($row=mysql_fetch_array($result)) {
        echo '<tr>';
        for ($i=0;$i<(count($row)/2);$i++) {
          echo '<td>',htmlentities($row[$i]),'</td>';
        }
        echo '</tr>';
      }
      echo '</table></p>';
    }
  }
return FALSE;
}
?>

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