те варианты скриптов, которые могут подключиться, считают что таблица пустая.
Пожалуйста, предложите рабочий код.
Мой код:
Код: Выделить всё
<html>
<head>
<title>Virtual Shop on PHP</title>
<meta charset="UTF-8">
<link rel='stylesheet' href='hakerstyle.css'/>
</head>
<body>
<a href="https://www.ho.ua">Хостинг В Украине бесплатно</a><br>
<?php
$servername = "db2.ho.ua";
$username = "sudilovskiy1onpu";
$password = "230598101";//for phpMyAdmin
echo "try 1<br>";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully<br>";
$sql = "SELECT * FROM firma";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
echo "id: " . $row["identifier"]. " - Name: " . $row["first_name"]. " " . $row["last_name"]." ".$row["status"]." ";
}
} else {
echo "0 results<br>";//почему я вижу эту строку?
}
echo "try 2<br>";
$set = mysql_query('SHOW DATABASES;');
$dbs = array();
while($db = mysql_fetch_row($set))
$dbs[] = $db[0];
echo implode('<br/>', $dbs);
//dosen't nothing
mysqli_close($conn);
echo "try 3<br>";
$con=mysqli_connect("db2.ho.ua","sudilovskiy1onpu","230598101","sudilovskiy1onpu");
// Check connection
if (mysqli_connect_errno()) echo "Failed to connect to MySQL: " . mysqli_connect_error();
$result = mysqli_query($con,"SELECT * FROM Persons");
echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['first_name'] . "</td>";
echo "<td>" . $row['last_name'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
</body>
</html>
try 1
Connected successfully
0 results
try 2
try 3
____________________
|Firstname|Lastname|
screenshots:
https://drive.google.com/open?id=1Un9Vw ... ZL2R2uedNe
https://drive.google.com/open?id=1MGclf ... IiAUE1hgsl