Question

我无法将随机天数小时和秒添加到当前日期。

这是我的代码 -

$randnos=rand(0,30);
$randhrs=rand(00,24);
$randmin=rand(00,60);
$randsec=rand(00,60);    

$newTime = date("d/m/Y H:m:s",strtotime(" +'.$randnos.' days" . " +'.$randhrs.' Hours". " +'.$randmin.' minutes". " +'.$randsec.' seconds"));

update_post_meta($post_id,'test',$newTime);

Answer 1

你的引号都搞砸了。

$randnos=rand(0,30);
$randhrs=rand(00,24);
$randmin=rand(00,60);
$randsec=rand(00,60);    

$newTime = date("d/m/Y H:m:s",strtotime("+$randnos days +$randhrs hours +$randmin minutes +$randsec seconds"));

update_post_meta($post_id,'test',$newTime);

Answer 2

<强>一衬垫

echo date('Y-m-d h:i:s', strtotime('+' .rand(30, 60 * 60 * 24 * 3).' seconds'));

Answer 3

public void ChercheStextBox_TextChanged(object sender, EventArgs e) { //NASSIM LOUCHANI BindingSource bs = new BindingSource(); bs.DataSource = dataGridView3.DataSource; bs.Filter = string.Format("CONVERT(" + dataGridView3.Columns[1].DataPropertyName + ", System.String) like '%" + ChercheStextBox.Text.Replace("'", "''") + "%'"); dataGridView3.DataSource = bs; }中使用的string中的问题。

strtotime()

首先使用随机数据创建字符串，然后插入$randnos=rand(0,30); $randhrs=rand(00,24); $randmin=rand(00,60); $randsec=rand(00,60); $str = ' +'.$randnos.' days +'.$randhrs.' Hours +'.$randmin.' minutes +'.$randsec.' seconds'; $newTime = date("d/m/Y H:m:s",strtotime($str));。

Answer 4

RewriteRule ^courses/([a-zA-Z0-9-_]+)/(0-9-]+)/?$ filename.php?param1=$1&param2=$2 [QSA,L]

无法在php中添加当前时间

4 个答案: