Question

我对这个php代码有真正的了解：

 <?php 
        $sessionid = $_SESSION['id'];
        echo $sessionid;

        $sql = "SELECT * FROM users WHERE id = '$sessionid';";
        $rsults = mysqli_query($conn, $sql);
        $resultsCheck = mysqli_num_rows($rsults);
        if ($resultsCheck > 0) {
            while ($row = mysqli_fetch_assoc($rsults)){
                $follow = $row['follow'];
                $loc = $row['places'];
                echo $follow;
                echo $loc;
            }
        }
    $sql = "SELECT * FROM posts WHERE username LIKE '$follow'";
    $rsults = mysqli_query($conn, $sql);
    $resultsCheck = mysqli_num_rows($rsults);
    if ($resultsCheck > 0) {
        while ($row = mysqli_fetch_assoc($rsults)){
            echo '<div class="posts">';
            echo '<img class="img"src='.$row['img'].' width="1500px">';
            echo '</div>';
            echo '<div class="contain">';
            echo '<div class="over">';
            echo '<div class="username2">';
            echo '<img src="focus.png" width="25px" height="25px" style="padding-right: 10px;">'.'<a href="./Profile.php?data='.$row['username'].'">'.$row['username'].'</a>'.'<img src="loc.png" width="25px" height="25px" style="padding-right: 5px; padding-left: 10px;">'.'<a href="./Location.php?data='.$row['ext'].'">'.$row['ext'].'</a>';
            echo '</div>';
            echo '<div class="content">';
            echo $row['content'];
            echo '</div>';         
            echo '</div>';
        }
    }else {
        echo "There are no results matching your search!";
}
?>

它将从第一个选择语句输出变量，但是第二个语句仅输出“没有符合您搜索条件的结果！”。我已经尝试在php我的管理员中使用实际的单词（不是变量）声明该语句，但是当它在站点上时，效果很好。我想这与变量有关，但实际上我不知道，因此任何帮助将不胜感激！非常感谢（而且我知道变量可以进行sql注入，但是此站点仅是概念证明：）

Answer 1

对于第二条SQL语句，您需要将%与LIKE运算符结合使用。 %-百分号代表零个，一个或多个字符

要更深入地了解，您可以点击以下链接：SQL LIKE Operator

更改您的第二条SQL语句，如下所示。我认为它会按您的预期工作。

$sql = "SELECT * FROM posts WHERE username LIKE '%$follow%'";

Answer 2

尝试一下，我做了一些调整，所以它可以工作

<?php 
        $sessionid = $_SESSION['id'];
        echo $sessionid;

        $sql = "SELECT * FROM users WHERE id = '".$sessionid."'";
        $rsults = mysqli_query($conn, $sql);
        $resultsCheck = mysqli_num_rows($rsults);
        if ($resultsCheck > 0) {
            while ($row = mysqli_fetch_array($rsults)){
                $follow = $row['follow'];
                $loc = $row['places'];
                echo $follow;
                echo $loc;
            }
        }
    $sql = "SELECT * FROM posts WHERE username LIKE '".$follow."'";
    $rsults = mysqli_query($conn, $sql);
    $resultsCheck = mysqli_num_rows($rsults);
    if ($resultsCheck > 0) {
        while ($row = mysqli_fetch_array($rsults)){
            echo '<div class="posts">';
            echo '<img class="img"src='.$row['img'].' width="1500px">';
            echo '</div>';
            echo '<div class="contain">';
            echo '<div class="over">';
            echo '<div class="username2">';
            echo '<img src="focus.png" width="25px" height="25px" style="padding-right: 10px;">'.'<a href="./Profile.php?data='.$row['username'].'">'.$row['username'].'</a>'.'<img src="loc.png" width="25px" height="25px" style="padding-right: 5px; padding-left: 10px;">'.'<a href="./Location.php?data='.$row['ext'].'">'.$row['ext'].'</a>';
            echo '</div>';
            echo '<div class="content">';
            echo $row['content'];
            echo '</div>';         
            echo '</div>';
        }
    }else {
        echo "There are no results matching your search!";
}
?>

PHP从语句中选择时，不起作用

2 个答案: