UPDATE更新查询的基础语法是
Where Update
查询的完成:
让我们斟酌下表“Data”,个中包括四列“ID”,“FirstName”,“LastName”和“Age”。
要更新“Data”表中“ID”为201的职员的“Age”,我们能够运用以下代码:
运用历程要领更新查询:
<?php $link = mysqli_connect("localhost", "root", "", "Mydb"); if($link === false){ die("ERROR: Could not connect. " . mysqli_connect_error()); } $sql = "UPDATE data SET Age='28' WHERE id=201"; if(mysqli_query($link, $sql)){ echo "Record was updated successfully."; } else { echo "ERROR: Could not able to execute $sql. " . mysqli_error($link); } mysqli_close($link); ?>
输出:更新后的表格
Web浏览器上的输出:
运用面向对象的要领更新查询:
<?php $mysqli = new mysqli("localhost", "root", "", "Mydb"); if($mysqli === false){ die("ERROR: Could not connect. " . $mysqli->connect_error); } $sql = "UPDATE data SET Age='28' WHERE id=201"; if($mysqli->query($sql) === true){ echo "Records was updated successfully."; } else{ echo "ERROR: Could not able to execute $sql. " . $mysqli->error; } $mysqli->close(); ?>
运用PDO要领更新查询:
<?php try{ $pdo = new PDO("mysql:host=localhost; dbname=Mydb", "root", ""); $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); } catch(PDOException $e){ die("ERROR: Could not connect. " . $e->getMessage()); } try{ $sql = "UPDATE data SET Age='28' WHERE id=201"; $pdo->exec($sql); echo "Records was updated successfully."; } catch(PDOException $e){ die("ERROR: Could not able to execute $sql. " . $e->getMessage()); } unset($pdo); ?>
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