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我对SQL很陌生,PHP想要在搜索数据库中显示图像以及其他结果,这可能吗?谁能为我提供任何帮助?如何使用PHP和SQL将图像添加到表中?
正如我所说,我对编码非常陌生,并且正在努力挣扎。
<?php
//load database connection
$host = "localhost";
$user = "root";
$password = "root";
$database_name = "catalog";
$pdo = new PDO("mysql:host=$host;dbname=$database_name", $user, $password,
array(
PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION
));
// Search from MySQL database table
$search=$_POST['search'];
$query = $pdo->prepare("select * from catalog.final_dog_catologue_full where
name LIKE '%$search%' OR Brand LIKE '%$search%' LIMIT 0 , 1000");
$query->bindValue(1, "%$search%", PDO::PARAM_STR);
$query->execute();
// Display search result
if (!$query->rowCount() == 0) {
echo "Search found :<br/>";
echo "<table style=\"font-family:arial;color:#333333;\">";
echo "<tr><td style=\"border-style:solid;border-
width:1px;border-color:#98bf21;background:#98bf21;\">Name</td><td
style=\"border-style:solid;border-width:1px;border-
color:#98bf21;background:#98bf21;\">Brand</td><td style=\"border-
style:solid;border-width:1px;border-
color:#98bf21;background:#98bf21;\">Price</td><td style=\"border-
style:solid;border-width:1px;border-
color:#98bf21;background:#98bf21;\">Category</td><td style=\"border-
style:solid;border-width:1px;border-
color:#98bf21;background:#98bf21;\">Animal</td></tr>";
while ($results = $query->fetch()) {
echo "<tr><td style=\"border-style:solid;border-
width:1px;border-color:#98bf21;\">";
echo $results['Name'];
echo "</td><td style=\"border-style:solid;border-
width:1px;border-color:#98bf21;\">";
echo $results['Brand'];
echo "</td><td style=\"border-style:solid;border-
width:1px;border-color:#98bf21;\">";
echo "£".$results['Retail_Price_With_Delievery'];
echo "</td><td style=\"border-style:solid;border-
width:1px;border-color:#98bf21;\">";
echo $results['Catogary'];
echo "</td><td style=\"border-style:solid;border-
width:1px;border-color:#98bf21;\">";
echo $results['Animal'];
echo "</td></tr>";
}
echo "</table>";
} else {
echo 'Nothing found';
}
?>
这是绝对可能的,因为你不是将图像**插入到数据库中,而是简单地从数据库中回显数据**。只需将图像与表格的其余部分一起回显出来;图像只是HTML。 –
正如我所说,我对编码非常陌生,并且在如何/在哪里添加创建“echo”的行来挣扎,这将允许我在显示的表中显示图片。 –