PHP mySQLi 更新表
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PHP mySQLi update table
提问by aandroidtest
Currently, I am using PHP to get some from the backend and insert into the database using mysqli. Below is the code used:
目前，我正在使用 PHP 从后端获取一些并使用 mysqli 插入到数据库中。下面是使用的代码：
$conn = new mysqli('localhost', 'username', 'pwd', 'db');
// check connection
if (mysqli_connect_errno()) {
exit('Connect failed: '. mysqli_connect_error());
}
$sql = "INSERT INTO `birthday` (`birthday`) VALUES ('$birthday')";
// Performs the $sql query and get the auto ID
if ($conn->query($sql) === TRUE) {
echo 'The auto ID is: '. $conn->insert_id;
}
else {
echo 'Error: '. $conn->error;
}
Now if I am going to fetch the information again, how to I update this value? Currently, it will create another row and insert the value again.
现在如果我要再次获取信息，我该如何更新这个值？目前，它将创建另一行并再次插入值。
Thanks In Advance
提前致谢
回答by OrganizedChaos
What I typically do is something like this.
我通常做的是这样的事情。
Also, you need to make sure you have a field or something that is unique to this record. Basically, it will always INSERT the way it's written, since we're just checking one value (birthday)
此外，您需要确保您有一个字段或该记录独有的内容。基本上，它总是按照它写的方式插入，因为我们只是检查一个值（生日）
Here's an example
这是一个例子
$conn = new mysqli('localhost', 'username', 'pwd', 'db');
// check connection
if (mysqli_connect_errno()) {
exit('Connect failed: '. mysqli_connect_error());
}
// check to see if the value you are entering is already there
$result = $conn->query("SELECT * FROM birthday WHERE name='Joe'");
if ($result->num_rows > 0){
// this person already has a b-day saved, update it
$conn->query("UPDATE birthday SET birthday = '$birthday' WHERE name = 'Joe'");
}else{
// this person is not in the DB, create a new ecord
$conn->query("INSERT INTO `birthday` (`birthday`,`name`) VALUES ('$birthday','Joe')");
}
