mysqli_fetch_array 循环问题

Question

<?php  
$q = isset($_POST["q"]) ? intval($_POST["q"]) : "";

if(empty($q)) {
    echo 'please choose a web site'; 
    exit;
}

$con = mysqli_connect('localhost','root','');

if(!$con) {
    die('Could not connect : '.mysqli_error($con));
}

//choose database
mysqli_select_db($con,'sixfive');

//set charset
mysqli_set_charset($con,'utf8');

$sql = "select * from websites where id='".$q."'";

$result = mysqli_query($con,$sql);

echo "<table border='1'>";
echo "<tr>";
echo "<th>ID</th>";
echo "<th>SiteName</th>";
echo "<th>SiteURL</th>";
echo "<th>ALex</th>";
echo "<th>Country</th>";
echo "</tr>";

while($row = mysqli_fetch_array($result)) {
    echo "<tr>";
    echo "<td>".$row['id']."</td>";
    echo "</tr>";
}

echo "</table>";

mysqli_close($con);

?>

$q是接受Ajax传过来的q的值

while($row = mysqli_fetch_array($result)) {
    echo "<tr>";
    echo "<td>".$row['id']."</td>";
    echo "</tr>";
}

这个while中的 条件判断不会死循环吗？ 运行结果是没有死循环 $row = mysqli_fetch_array($result) 返回的值不是一直是true吗？但是运行结果正常 哪里的问题呢？

Answer 1

mysqli_fetch_array这个函数是从结果集中取一行返回，如果没有更多的行则返回false，所以不存在死循环这个操作的