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PHP语法错误,几乎没有帮助

[英]Php syntax error, little help

I have this string: 我有这个字符串:

$imagename ="$ad_id_stripped"."_1".".jpg";
$display_table.="<a href='../ad.php?ad_id=$row[ad_id]' target='_parent'>
<img style='border:none;' src='../ad_images/$category/thumbs/$imagename?time()' class='shadow'></a>";

echo $display_table;

As you maybe can see I am trying to add the time() function in there... However, there is no time() added to it like this! 如您可能看到的那样,我正在尝试在其中添加time()函数。但是,没有像这样向其中添加time()! I have tried with various quotes etc without luck... 我尝试过各种报价等,没有运气...

Any ideas? 有任何想法吗?

Thanks 谢谢

$imagename ="$ad_id_stripped"."_1".".jpg";
$display_table.="<a href='../ad.php?ad_id=$row[ad_id]' target='_parent'>
<img style='border:none;' src='../ad_images/$category/thumbs/$imagename?" . time() . "' class='shadow'></a>";

echo $display_table;

You want to get the time() function out of the string itself and concatenate 您想从字符串本身中获取time()函数并进行连接

You can't do: 您不能:

$string = "....time()...";

You will need to do: 您需要执行以下操作:

$string = "..." . time() . "....";

or 要么

$time = time();
$string = "....$time...";

Strings wrapped in double quotes will parse variables, but not functions. 用双引号引起来的字符串将解析变量,但不解析函数。 Try this: 尝试这个:

$imagename ="$ad_id_stripped"."_1".".jpg";
$time = time();
$display_table.="<a href='../ad.php?ad_id=$row[ad_id]' target='_parent'>
<img style='border:none;' src='../ad_images/$category/thumbs/$imagename?$time' class='shadow'></a>";

echo $display_table;

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