我试图通过仅获取其名称的一部分来读取存储在 PHP 目录中的图像。图像命名如下,
AT-1410f1654.jpg
AT-1410_1655.jpg
AT-1410_1656.jpg
AT-1410_1657.jpg
AT-1410_1658.jpg
AT-1410_1659.jpg
我已经尝试了下面的代码,但即使我使用 PHP substr() 方法它也不起作用,但它也不起作用,
$dbImage=$row["pref"];
$imageName=$dbImage;
$extension=".jpg";
$filename=$imageName.$extension;
echo "$filename+<img src='proppics/$filename'>";
关于如何做到这一点的任何想法
完整代码
$limit=10;
$con=mysql_connect("localhost","root","");
mysql_select_db("movedb") or die("Unable to select database");
$query="select * FROM properties where `name`='Beata Grande 1' & `catergory`='Villas'& `price`=202800 & `area`='Arenas'& `bedrooms`=2 & `region`='Axarquia'";
$numresults=mysql_query($query,$con);
$numrows=mysql_num_rows($numresults);
$result = mysql_query($query) or die("Couldn't execute query");
echo "<center>";
echo "<p>You searched for: "" . $properties . ""</p>";
echo "<form name=payment action='properties_details.php'>";
echo "Results <br>";
while ($row= mysql_fetch_array($result)) {
$id=$row['id'];
$pid=$row['pref'];
// Retrieve the balance database fields
echo "<p>Property ID  ".$pid;
echo "<br> <p> Name  ";
echo $row["name"];
echo "<br> Properties  ";
echo $row["catergory"];
echo "<br> Description  ";
// Print results
echo "<br>";
echo "<input type=submit name=btnbuy value=MoreDetails> ";