Question

我有这个简单的搜索表单，但没有返回任何内容。当我尝试var_dump($ser)时显示null。知道为什么会这样吗？

include 'include/db.php';

$sql = mysqli_query($conn, "SELECT caption,name FROM images WHERE caption LIKE '%$_GET[caption]%' LIMIT 0,1");

$caption = $sql['caption'];
$name = $sql['name'];

while($ser = mysqli_fetch_array($sql)){

$out .= "<div id=\"picture\">";
$out .= "<img style=\"width:100%;margin:0 auto;\" src=\"/upload/".$ser['name']."\" /><br />";
$out .= "<div id=\"caption\">";
$out .= "<h1>" . $ser['caption'] . "</h1>";
$out .= "</div><br />";
}

这是表格

    <form action="../search.php" method="get">
    <input type="text" name="caption" size="15">
    <input type="submit" value="Search">
</form>

非常基本，但仍然很难理解错误。

Answer 1

请试试这个

$result= mysqli_query($conn, "select caption,name from images where caption like '%'") or die (mysqli_error());

while($line = mysqli_fetch_array($result))
{
    echo $line['caption'].$line['name']."<br/>";
}

检查您是否有错误或某些结果

Answer 2

$caption = $_GET['caption'];

$sql = mysqli_query($conn, "SELECT caption,name FROM images WHERE caption LIKE '%'".$caption."'%' LIMIT 0,1");

while($ser = mysqli_fetch_array($sql)){

   echo $ser['name']." ".$ser['caption'];
}

简单的搜索表单不会返回任何结果

2 个答案: