PHP的getimagesize与变量

本文介绍了PHP的getimagesize与变量的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我正在尝试使用getimagesize函数获取图像的高度和高度.我正在从数据库中提取图像URL. (字段ProjectURL包含一行，例如xxx.jpg).但是我遇到了错误.

I'm trying to use the getimagesize function to get the height and with of an image. I'm pulling the image URL from a database. (The field ProjectURL contains a line such as xxx.jpg). However I'm getting an error.

代码:

$testing = "projects/'.$row['ProjectURL'].'";
    list($width, $height, $type, $attr) = getimagesize($testing);
    echo "Image width " .$width;
echo "<br />";
echo "Image height " .$height;

错误:

推荐答案

这是因为您混合使用单引号和双引号...

it's because you are mixing single and double quotes...

这应该没问题:

$testing = "projects/" . $row['ProjectURL'];
list($width, $height, $type, $attr) = getimagesize($testing);
echo "Image width " . $width;
echo "Image height " . $height;

您可能还已经注意到，我删除了 echo"; ...这个没用了:)

You might also have noticed that I removed the echo "";... this one was useless :)

这篇关于PHP的getimagesize与变量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！