当我尝试这段代码时,我收到关于第二个参数到字符串的警告。我在之前类似的问题中看到过一些答案,但我没有找到解决方案…据我所知,问题是if语句??谢谢。
if (isset($_GET['id'])) {
$str_id = $_GET['id'];
($conn->set_charset("utf8"));
if ($result=mysqli_query($conn, $q )) {
while ($obj=mysqli_fetch_object($result)) {
}
?>
<?php
$con=mysqli_connect("localhost","my_user","my_password","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="SELECT Lastname,Age FROM Persons ORDER BY Lastname";
if ($result=mysqli_query($con,$sql))
{
while ($obj=mysqli_fetch_object($result))
{
printf("%s (%s)'n",$obj->Lastname,$obj->Age);
}
// Free result set
mysqli_free_result($result);
}
mysqli_close($con);
?>