如果 ID 已经存在，我想更新我的数据库；如果 ID 不存在，我想插入。
process.php:
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "maindata";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("<center>Connection failed: " . mysqli_connect_error() . "</center");
}
$id = $_POST['id'];
$aname = $_POST['aname'];
$xsummary = $_POST['xsummary'];
$sql=mysql_query("SELECT * FROM info WHERE ID = $id");
if (mysql_num_rows($sql) == $id)
{
$sql="update info set AccountName= '$aname', ExecutiveSummary='$xsummary'";
}
else {
$sql="insert into AccountName= '$aname', ExecutiveSummary= '$xsummary'";
}
有人帮我解决了我的问题。
提前致谢。
请看下面的代码。
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "maindata";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("<center>Connection failed: " . mysqli_connect_error() . "</center");
}
$id = $_POST['id'];
$aname = $_POST['aname'];
$xsummary = $_POST['xsummary'];
$sql=mysqli_query($conn, "SELECT * FROM info WHERE ID = $id");
if (mysqli_num_rows($sql) > 0)
{
$sqlUpdate="update info set AccountName= '$aname', ExecutiveSummary= '$xsummary'";
mysqli_query($conn,$sqlUpdate);
} else {
$sqlInsert="insert into info set AccountName= '$aname', ExecutiveSummary= '$xsummary'";
mysqli_query($conn,$sqlInsert);
}
您在代码中的实际问题，您是使用 mysqli 连接数据库，但使用 mysql 从表中获取数据。请使用此代码，它将正常工作。
*注意:- 插入语句中存在错误，请更新，这将起作用。