MikeW提出了一个有趣的解决方案，但正如评论中提到的，它是一个SQL Server解决方案而不是MySQL解决方案。下面是一个MySQL解决方案:

use information_schema;
set @q = 'Boston';
set @t = 'my_db';

select CONCAT('use \'',@q,'\';') as q UNION
select CONCAT('select \'', tbl.`TABLE_NAME`,'\' as TableName, \'', col.`COLUMN_NAME`,'\' as Col, `',col.`COLUMN_NAME`,'` as value  from `' , tbl.`TABLE_NAME`,'` where `' ,
        col.`COLUMN_NAME` , '` like \'%' ,@q,  '%\' UNION') AS q
from `tables` tbl
inner join `columns` col on tbl.`TABLE_NAME` = col.`TABLE_NAME`and col.DATA_TYPE='varchar'
where tbl.TABLE_SCHEMA  =  @t  ;

这里有两个简单的代码。首先执行mysqldump:

mysqldump -uUSERNAME -p DATABASE_NAME > database-dump.sql

然后grep sqldump文件:

grep -i "Search string" database-dump.sql

也可以找到/替换并重新导入到数据库中。

如果你可以使用bash -这里有一个脚本: 它需要一个用户dbread，并在数据库上传递dbread。

#!/bin/bash
IFS='
'
DBUSER=dbread
DBPASS=dbread
echo -n "Which database do you want to search in (press 0 to see all databases): " 
read DB
echo -n "Which string do you want to search: " 
read SEARCHSTRING
for i in `mysql $DB -u$DBUSER -p$DBPASS -e "show tables" | grep -v \`mysql $DB -u$DBUSER -p$DBPASS -e "show tables" | head -1\``
do
for k in `mysql $DB -u$DBUSER -p$DBPASS -e "desc $i" | grep -v \`mysql $DB -u$DBUSER -p$DBPASS -e "desc $i" | head -1\` | grep -v int | awk '{print $1}'`
do
if [ `mysql $DB -u$DBUSER -p$DBPASS -e "Select * from $i where $k='$SEARCHSTRING'" | wc -l` -gt 1 ]
then
echo " Your searchstring was found in table $i, column $k"
fi
done
done

如果有人想要一个解释:http://infofreund.de/?p=1670

当我们在Wordpress网站上更改域名时，我自己也在寻找这个。没有编程是做不到的，这就是我所做的。

<?php  
  header("Content-Type: text/plain");

  $host = "localhost";
  $username = "root";
  $password = "";
  $database = "mydatabase";
  $string_to_replace  = 'old.example.com';
  $new_string = 'new.example.com';

  // Connect to database server
  mysql_connect($host, $username, $password);

  // Select database
  mysql_select_db($database);

  // List all tables in database
  $sql = "SHOW TABLES FROM ".$database;
  $tables_result = mysql_query($sql);

  if (!$tables_result) {
    echo "Database error, could not list tables\nMySQL error: " . mysql_error();
    exit;
  }

  echo "In these fields '$string_to_replace' have been replaced with '$new_string'\n\n";
  while ($table = mysql_fetch_row($tables_result)) {
    echo "Table: {$table[0]}\n";
    $fields_result = mysql_query("SHOW COLUMNS FROM ".$table[0]);
    if (!$fields_result) {
      echo 'Could not run query: ' . mysql_error();
      exit;
    }
    if (mysql_num_rows($fields_result) > 0) {
      while ($field = mysql_fetch_assoc($fields_result)) {
        if (stripos($field['Type'], "VARCHAR") !== false || stripos($field['Type'], "TEXT") !== false) {
          echo "  ".$field['Field']."\n";
          $sql = "UPDATE ".$table[0]." SET ".$field['Field']." = replace(".$field['Field'].", '$string_to_replace', '$new_string')";
          mysql_query($sql);
        }
      }
      echo "\n";
    }
  }

  mysql_free_result($tables_result);  
?>

希望它能帮助任何在未来遇到这个问题的人:)

一个简单的解决方案是这样做的:

mysqldump -u myuser --no-create-info --extended-insert=FALSE databasename | grep -i "<search string>"

本视频的前30秒展示了如何使用Phpmyadmin的全局搜索功能 这很有效。它将在每个表中搜索字符串。

http://www.vodahost.com/vodatalk/phpmyadmin-setup/62422-search-database-phpmyadmin.html