PHP getimagesize变量

Question

我正在尝试使用该getimagesize函数来获取高度和图像.我从数据库中提取图像URL.(该字段ProjectURL包含一行等xxx.jpg).但是我收到了一个错误.

码:

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

错误:

解析错误:语法错误,意外T_ENCAPSED_AND_WHITESPACE,期待T_STRING或T_VARIABLE或T_NUM_STRING

Answer 1

这是因为你混合了单引号和双引号......

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;

你可能也注意到我删除了回音""; ......这一个没用:)