php mysqli未返回值

下面是给我带来问题的代码：

$letter = $_POST['artistButton'];
$result = $mysqli->query("SELECT * FROM `Generic` WHERE `songName` LIKE '$letter%'");
$row    = $result->fetch_array();

while ($row = $result->fetch_array()) {
    $Artist = $row['Artist'];
    $songName = $row['songName'];
    $Duration = $row['Duration'];
    $URL = $row['URL'];
    $Genre = $row['Genre'];
    echo "<input type='submit' value='$songName' name='artistButton'>";
    echo "<br />";
}

$letter=$\u POST['artistButton']；
$result=$mysqli->query（“从`Generic`WHERE`songName`LIKE'$letter%'中选择*”；
$row=$result->fetch_array（）；
而（$row=$result->fetch_array（））{
$Artist=$row['Artist']；
$songName=$row['songName']；
$Duration=$row['Duration']；
$URL=$row['URL']；
$GREEP=$row['GREEP']；
回声“；
回声“
”；
}

---

语句是正确的我将它保存到一个变量并打印到屏幕上，然后从数据库运行它，它打印了我想要的结果，所以它肯定不是这样。由于某种原因，尽管它没有显示。我猜这与我把mysql转换成mysqli的过程搞砸了有关。谢谢-Sam

显然有些代码被删除了，但请尝试删除额外的代码：

$letter = $_POST['artistButton'];
$result = $mysqli->query("SELECT * FROM `Generic` WHERE `songName` LIKE '$letter%'");
$row    = $result->fetch_array();

while ($row = $result->fetch_array()) {
    $Artist = $row['Artist'];
    $songName = $row['songName'];
    $Duration = $row['Duration'];
    $URL = $row['URL'];
    $Genre = $row['Genre'];
    echo "<input type='submit' value='$songName' name='artistButton'>";
    echo "<br />";
}

$row = $result->fetch_array();

和使用：

$letter = $_POST['artistButton'];
$result = $mysqli->query("SELECT * FROM `Generic` WHERE `songName` LIKE '$letter%'");
$row    = $result->fetch_array();

while ($row = $result->fetch_array()) {
    $Artist = $row['Artist'];
    $songName = $row['songName'];
    $Duration = $row['Duration'];
    $URL = $row['URL'];
    $Genre = $row['Genre'];
    echo "<input type='submit' value='$songName' name='artistButton'>";
    echo "<br />";
}

while($row = $result->fetch_assoc()) {

因为你对关联索引感兴趣

$letter = $_POST['artistButton'];
$result = $mysqli->query("SELECT * FROM `Generic` WHERE `songName` LIKE '$letter%'");
$row    = $result->fetch_array();

while ($row = $result->fetch_array()) {
    $Artist = $row['Artist'];
    $songName = $row['songName'];
    $Duration = $row['Duration'];
    $URL = $row['URL'];
    $Genre = $row['Genre'];
    echo "<input type='submit' value='$songName' name='artistButton'>";
    echo "<br />";
}

重要提示：由于您使用的是

mysqli*

，因此请改用它的参数化查询。不要直接在查询上使用

$\u POST

值

$letter = $_POST['artistButton'];
$result = $mysqli->query("SELECT * FROM `Generic` WHERE `songName` LIKE '$letter%'");
$row    = $result->fetch_array();

while ($row = $result->fetch_array()) {
    $Artist = $row['Artist'];
    $songName = $row['songName'];
    $Duration = $row['Duration'];
    $URL = $row['URL'];
    $Genre = $row['Genre'];
    echo "<input type='submit' value='$songName' name='artistButton'>";
    echo "<br />";
}

$letter = "{$_POST[artistButton]}%";
$stmt = $mysqli->prepare('SELECT * FROM `Generic` WHERE `songName` LIKE ?');
$stmt->bind_param('s', $letter);
$stmt->execute();

$result = $stmt->get_result();

while($row = $result->fetch_assoc()) {
    $Artist = $row['Artist'];
    $songName = $row['songName'];
    $Duration = $row['Duration'];
    $URL = $row['URL'];
    $Genre = $row['Genre'];
    echo "<input type='submit' value='$songName' name='artistButton'>";
    echo "<br />";
}

$letter=“{$\u POST[artistButton]}%”；
$stmt=$mysqli->prepare（'SELECT*FROM`Generic`WHERE`songName`LIKE？'）；
$stmt->bind_参数（'s'，$letter）；
$stmt->execute（）；
$result=$stmt->get_result（）；
而（$row=$result->fetch_assoc（））{
$Artist=$row['Artist']；
$songName=$row['songName']；
$Duration=$row['Duration']；
$URL=$row['URL']；
$GREEP=$row['GREEP']；
回声“；
回声“
”；
}

删除多余的

$row=$result->fetch_array（）在
$result
下面，并在MYSQLI中再次尝试。没有准备好的语句，您开始得很好。这被我拖得很慢，但仍然不起作用，因为您与数据库有连接？并在
fetch_array（）
=
MYSQLI_ASSOC
中添加一个参数，您确定它位于前导的
%
？不是
%$letter%
，提示：使用预先准备好的语句，不管怎样，它都是mysqli