页: 1
1. "%Y%d%m-text.TXT"
2. "20112104-text.TXT"
我如何利用选定格式获得<代码>20112104”? 我可以采用第一种显示格式,%d%d”,我的产出应为212121。 事先感谢你。
你们正在寻求的是吗?
$sFile = "%s-text.TXT";
$sFile = sprintf($sFile, date("Ydm"));
如果没有,请补充更多内容。
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