警告:mysqli_query() 需要至少 2 个参数,其中 1 个在我的 php 文件中给出
Warning: mysqli_query() expects at least 2 parameters, 1 given in in my php file
我正在尝试在 php 代码中编写代码以更新 table 但是我在这里收到警告为什么我在这里收到此警告
这是我的代码
$con=mysqli_connect("localhost","root","root","clothing");
// Check connection
if (mysqli_connect_errno()){
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if(isset($_POST['change'])){
$email=$_POST['email'];
$sql="SELECT *from userdetails WHERE userId='$userid'";
$query = mysqli_query($con, $sql);
$count=mysqli_num_rows($query);
$row=mysqli_fetch_array($query);
if ($count > 0){
echo $row["Password"];
echo $_POST["currentPassword"];
if($_POST["currentPassword"] == $row["Password"]) {
echo "fine";
if( $_POST["confirmPassword"]== $_POST["newPassword"]){
mysqli_query("UPDATE userdetails SET Password='anil' WHERE ID=30");
echo "good";
}
}
}
我怎样才能得到我想要的输出
提前致谢
您没有向 mysqli_query - $con 提供这两个参数。使用下面的代码
$con=mysqli_connect("localhost","root","root","clothing");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if(isset($_POST['change']))
{
$email=$_POST['email'];
$sql="SELECT *from userdetails WHERE userId='$userid'";
$query = mysqli_query($con, $sql);
$count=mysqli_num_rows($con,$query);
$row=mysqli_fetch_array($con,$query);
if ($count > 0){
echo $row["Password"];
echo $_POST["currentPassword"];
if($_POST["currentPassword"] == $row["Password"]) {
echo "fine";
if( $_POST["confirmPassword"]== $_POST["newPassword"])
{
mysqli_query($con,"UPDATE userdetails SET Password='anil' WHERE ID=30");
echo "good";
}
}
}
}
希望对您有所帮助
您没有在查询中包含 $con。
mysqli_query($con, "UPDATE userdetails SET Password='anil' WHERE ID=30");
如果不添加 $con,您的查询将失败。
我正在尝试在 php 代码中编写代码以更新 table 但是我在这里收到警告为什么我在这里收到此警告
这是我的代码
$con=mysqli_connect("localhost","root","root","clothing");
// Check connection
if (mysqli_connect_errno()){
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if(isset($_POST['change'])){
$email=$_POST['email'];
$sql="SELECT *from userdetails WHERE userId='$userid'";
$query = mysqli_query($con, $sql);
$count=mysqli_num_rows($query);
$row=mysqli_fetch_array($query);
if ($count > 0){
echo $row["Password"];
echo $_POST["currentPassword"];
if($_POST["currentPassword"] == $row["Password"]) {
echo "fine";
if( $_POST["confirmPassword"]== $_POST["newPassword"]){
mysqli_query("UPDATE userdetails SET Password='anil' WHERE ID=30");
echo "good";
}
}
}
我怎样才能得到我想要的输出
提前致谢
您没有向 mysqli_query - $con 提供这两个参数。使用下面的代码
$con=mysqli_connect("localhost","root","root","clothing");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if(isset($_POST['change']))
{
$email=$_POST['email'];
$sql="SELECT *from userdetails WHERE userId='$userid'";
$query = mysqli_query($con, $sql);
$count=mysqli_num_rows($con,$query);
$row=mysqli_fetch_array($con,$query);
if ($count > 0){
echo $row["Password"];
echo $_POST["currentPassword"];
if($_POST["currentPassword"] == $row["Password"]) {
echo "fine";
if( $_POST["confirmPassword"]== $_POST["newPassword"])
{
mysqli_query($con,"UPDATE userdetails SET Password='anil' WHERE ID=30");
echo "good";
}
}
}
}
希望对您有所帮助
您没有在查询中包含 $con。
mysqli_query($con, "UPDATE userdetails SET Password='anil' WHERE ID=30");
如果不添加 $con,您的查询将失败。