PHP如何操作sqlite数据库？

连接数据库

下面的 PHP 代码显示了如何连接到一个现有的数据库。如果数据库不存在，那么它就会被创建，最后将返回一个数据库对象。

<?php
   class MyDB extends SQLite3
   {
      function __construct()
      {
         $this->open('test.db');
      }
   }
   $db = new MyDB();
   if(!$db){
      echo $db->lastErrorMsg();
   } else {
      echo "Opened database successfully\n";
   }
?>

现在，让我们来运行上面的程序，在当前目录中创建我们的数据库 test.db。您可以根据需要改变路径。如果数据库成功创建，那么会显示下面所示的消息：

Open database successfully

创建表

下面的 PHP 代码段将用于在先前创建的数据库中创建一个表：

<?php
   class MyDB extends SQLite3
   {
      function __construct()
      {
         $this->open('test.db');
      }
   }
   $db = new MyDB();
   if(!$db){
      echo $db->lastErrorMsg();
   } else {
      echo "Opened database successfully\n";
   }

   $sql =<<<EOF
      CREATE TABLE COMPANY
      (ID INT PRIMARY KEY     NOT NULL,
      NAME           TEXT    NOT NULL,
      AGE            INT     NOT NULL,
      ADDRESS        CHAR(50),
      SALARY         REAL);
EOF;

   $ret = $db->exec($sql);
   if(!$ret){
      echo $db->lastErrorMsg();
   } else {
      echo "Table created successfully\n";
   }
   $db->close();
?>

上述程序执行时，它会在 test.db 中创建 COMPANY 表，并显示下面所示的消息：

Opened database successfully
Table created successfully

INSERT 操作

下面的 PHP 程序显示了如何在上面创建的 COMPANY 表中创建记录：

<?php
   class MyDB extends SQLite3
   {
      function __construct()
      {
         $this->open('test.db');
      }
   }
   $db = new MyDB();
   if(!$db){
      echo $db->lastErrorMsg();
   } else {
      echo "Opened database successfully\n";
   }

   $sql =<<<EOF
      INSERT INTO COMPANY (ID,NAME,AGE,ADDRESS,SALARY)
      VALUES (1, 'Paul', 32, 'California', 20000.00 );

      INSERT INTO COMPANY (ID,NAME,AGE,ADDRESS,SALARY)
      VALUES (2, 'Allen', 25, 'Texas', 15000.00 );

      INSERT INTO COMPANY (ID,NAME,AGE,ADDRESS,SALARY)
      VALUES (3, 'Teddy', 23, 'Norway', 20000.00 );

      INSERT INTO COMPANY (ID,NAME,AGE,ADDRESS,SALARY)
      VALUES (4, 'Mark', 25, 'Rich-Mond ', 65000.00 );
EOF;

   $ret = $db->exec($sql);
   if(!$ret){
      echo $db->lastErrorMsg();
   } else {
      echo "Records created successfully\n";
   }
   $db->close();
?>

上述程序执行时，它会在 COMPANY 表中创建给定记录，并会显示以下两行：

Opened database successfully
Records created successfully

SELECT 操作

下面的 PHP 程序显示了如何从前面创建的 COMPANY 表中获取并显示记录：

<?php
   class MyDB extends SQLite3
   {
      function __construct()
      {
         $this->open('test.db');
      }
   }
   $db = new MyDB();
   if(!$db){
      echo $db->lastErrorMsg();
   } else {
      echo "Opened database successfully\n";
   }

   $sql =<<<EOF
      SELECT * from COMPANY;
EOF;

   $ret = $db->query($sql);
   while($row = $ret->fetchArray(SQLITE3_ASSOC) ){
      echo "ID = ". $row['ID'] . "\n";
      echo "NAME = ". $row['NAME'] ."\n";
      echo "ADDRESS = ". $row['ADDRESS'] ."\n";
      echo "SALARY =  ".$row['SALARY'] ."\n\n";
   }
   echo "Operation done successfully\n";
   $db->close();
?>

上述程序执行时，它会产生以下结果：

Opened database successfully
ID = 1
NAME = Paul
ADDRESS = California
SALARY =  20000

ID = 2
NAME = Allen
ADDRESS = Texas
SALARY =  15000

ID = 3
NAME = Teddy
ADDRESS = Norway
SALARY =  20000

ID = 4
NAME = Mark
ADDRESS = Rich-Mond
SALARY =  65000

Operation done successfully

UPDATE 操作

下面的 PHP 代码显示了如何使用 UPDATE 语句来更新任何记录，然后从 COMPANY 表中获取并显示更新的记录：

<?php
   class MyDB extends SQLite3
   {
      function __construct()
      {
         $this->open('test.db');
      }
   }
   $db = new MyDB();
   if(!$db){
      echo $db->lastErrorMsg();
   } else {
      echo "Opened database successfully\n";
   }
   $sql =<<<EOF
      UPDATE COMPANY set SALARY = 25000.00 where ID=1;
EOF;
   $ret = $db->exec($sql);
   if(!$ret){
      echo $db->lastErrorMsg();
   } else {
      echo $db->changes(), " Record updated successfully\n";
   }

   $sql =<<<EOF
      SELECT * from COMPANY;
EOF;
   $ret = $db->query($sql);
   while($row = $ret->fetchArray(SQLITE3_ASSOC) ){
      echo "ID = ". $row['ID'] . "\n";
      echo "NAME = ". $row['NAME'] ."\n";
      echo "ADDRESS = ". $row['ADDRESS'] ."\n";
      echo "SALARY =  ".$row['SALARY'] ."\n\n";
   }
   echo "Operation done successfully\n";
   $db->close();
?>

上述程序执行时，它会产生以下结果：

Opened database successfully
1 Record updated successfully
ID = 1
NAME = Paul
ADDRESS = California
SALARY =  25000

ID = 2
NAME = Allen
ADDRESS = Texas
SALARY =  15000

ID = 3
NAME = Teddy
ADDRESS = Norway
SALARY =  20000

ID = 4
NAME = Mark
ADDRESS = Rich-Mond
SALARY =  65000

Operation done successfully

DELETE 操作

下面的 PHP 代码显示了如何使用 DELETE 语句删除任何记录，然后从 COMPANY 表中获取并显示剩余的记录：

<?php
   class MyDB extends SQLite3
   {
      function __construct()
      {
         $this->open('test.db');
      }
   }
   $db = new MyDB();
   if(!$db){
      echo $db->lastErrorMsg();
   } else {
      echo "Opened database successfully\n";
   }
   $sql =<<<EOF
      DELETE from COMPANY where ID=2;
EOF;
   $ret = $db->exec($sql);
   if(!$ret){
     echo $db->lastErrorMsg();
   } else {
      echo $db->changes(), " Record deleted successfully\n";
   }

   $sql =<<<EOF
      SELECT * from COMPANY;
EOF;
   $ret = $db->query($sql);
   while($row = $ret->fetchArray(SQLITE3_ASSOC) ){
      echo "ID = ". $row['ID'] . "\n";
      echo "NAME = ". $row['NAME'] ."\n";
      echo "ADDRESS = ". $row['ADDRESS'] ."\n";
      echo "SALARY =  ".$row['SALARY'] ."\n\n";
   }
   echo "Operation done successfully\n";
   $db->close();
?>

上述程序执行时，它会产生以下结果：

Opened database successfully
1 Record deleted successfully
ID = 1
NAME = Paul
ADDRESS = California
SALARY =  25000

ID = 3
NAME = Teddy
ADDRESS = Norway
SALARY =  20000

ID = 4
NAME = Mark
ADDRESS = Rich-Mond
SALARY =  65000

Operation done successfully