Warning: file_get_contents(/data/phpspider/zhask/data//catemap/1/php/227.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
php mysqli获取数组未输出结果_Php_Syntax Error - Fatal编程技术网

php mysqli获取数组未输出结果

php mysqli获取数组未输出结果,php,syntax-error,Php,Syntax Error,所以我是mysqli的新手,我正在尝试从数据库中的一个表中回显结果 <?php $link = mysqli_connect("localhost", "root", "", "juneausmashbros"); if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit(); } $query = "SELECT * FROM post"; $r

所以我是mysqli的新手,我正在尝试从数据库中的一个表中回显结果

<?php
$link = mysqli_connect("localhost", "root", "", "juneausmashbros");

if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

$query = "SELECT * FROM post";
$result = mysqli_query($link, $query);

while($row =  $result->fetch_array()); {
    echo '<div><b>';
    echo $row['title'];
    echo '</b></div>';
    echo "\n";
}?> 

愚蠢的额外分号

while($row =  $result->fetch_array()); {
                                     ^

你有什么错误吗