[英]PHP getimagesize with variable
I'm trying to use the getimagesize
function to get the height and with of an image. 我正在尝试使用
getimagesize
函数来获取图像的高度和图像。 I'm pulling the image URL from a database. 我正在从数据库中提取图像URL。 (The field
ProjectURL
contains a line such as xxx.jpg
). (字段
ProjectURL
包含一行,例如xxx.jpg
)。 However I'm getting an error. 但是我遇到一个错误。
Code: 码:
$testing = "projects/'.$row['ProjectURL'].'";
list($width, $height, $type, $attr) = getimagesize($testing);
echo "Image width " .$width;
echo "<br />";
echo "Image height " .$height;
Error: 错误:
Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING
解析错误:语法错误,意外的T_ENCAPSED_AND_WHITESPACE,预期为T_STRING或T_VARIABLE或T_NUM_STRING
it's because you are mixing single and double quotes... 这是因为您混合使用单引号和双引号...
this should be ok: 这应该没关系:
$testing = "projects/" . $row['ProjectURL'];
list($width, $height, $type, $attr) = getimagesize($testing);
echo "Image width " . $width;
echo "Image height " . $height;
You might also have noticed that I removed the echo ""; 您可能还注意到我删除了回声“”; ... this one was useless :)
...这没用:)
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