Question

我正在使用$ _POST将数据发送到我的php页面来更新mysql数据库。我正在尝试使用客户ID来选择行并更新公司名称和fname（名字）。当我弄清楚这一点时，我将添加其余内容进行更新。我还通过“//”谢谢你

包括我尝试过的内容

-----dbconnect-----
$id= $_POST['id'];
$company= $_POST['company'];
$fname = $_POST['fname'];

echo $id;
echo $company;
echo $fname;
//$sql = mysqli_query($con,"UPDATE customer SET company = $company WHERE id= '.$id.'")
//$sql = "UPDATE customer SET company ='".mysql_real_escape_string($_POST['company'])."WHERE id='".mysql_real_escape_string($_POST['id'])."'";
$sql = "UPDATE customer SET company = $company WHERE id= '1'";
mysqli_select_db('customer');
$retval = mysqli_query( $sql, $con );
if(! $retval )
{
die('Could not update data: ' . mysqli_error());
}
echo "Updated data successfully\n";
mysqli_close($conn);
}

Answer 1

在这里做一些假设，但试试这个...

// make mysqli throw exceptions on error
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

// assuming your connection looks something like this
// you can pass the database name as the fourth argument instead of using select_db()
$con = new mysqli('localhost', 'user', 'pass', 'customer');

// $id = $_POST['id'], etc

// use a prepared statement with parameter placeholders.
// for more info see http://php.net/manual/mysqli.quickstart.prepared-statements.php
$stmt = $con->prepare('UPDATE customer SET company = ? WHERE id = ?');

// bind parameter variables and execute
$stmt->bind_param('si', $company, $id);
$stmt->execute();

echo 'Updated data successfully', PHP_EOL;

mysql更新无法正常工作..需要更新代码的帮助

1 个答案: