似乎有问题。当第52行if (mysqli_num_rows{$search_Result})
给我解析错误:语法错误,意外。
<?php
$host = "localhost";
$user = "root";
$password = "";
$database ="zebase";
$id="";
$name="";
$address="";
$actnum="";
$status="";
$remarks="";
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
function getPosts()
{
$Post = array();
$POST[0]= $_POST['id'];
$POST[1]= $_POST['name'];
$POST[2]= $_POST['address'];
$POST[3]= $_POST['actnum'];
$POST[4]= $_POST['status'];
$POST[5]= $_POST['remarks'];
return $posts;
}
if(isset($_POST['search']))
{
$data = getPosts();
$search_Query = "SELECT * FROM thetable WHERE id = $data[o]";
$search_Result = mysqli_query($conn, $search_Query);
}
if ($search_Result)
{
if (mysqli_num_rows{$search_Result})
{
while($row = mysqli_fetch_array($search_Result)}
{
$id =$row['id'];
$name =$row['name'];
$address =$row['address'];
$status =$row['status'];
$remarks =$row['remarks'];
}else {
echo 'No Data For This Id');
}
}else {
echo 'Result Error';
}
}
?>
答案 0 :(得分:0)
我认为您需要将该行更改为
如果(mysqli_num_rows($ search_Result))
如果希望可以解决该错误