在字符串中某个字符位置之前搜索值。

Question

我正在制作一个显示备份信息的网站。我遇到的障碍是我需要在某个角色位置之前获取信息。字符串看起来像：数据 1 => 2019-06-30-08-00-00 , 2019-07-30-08-00-00 , 2019-08-30-08-00-00 数据 2 => 2019-06-30-08-30 -00 , 2019-07-30-08-30-00 , 2019-08-30-08-30-00我从字符串中获取所有搜索信息的代码。$BackupJob_list = $BackupJob_list['Data'];$BackupJob_list = json_encode($BackupJob_list);$string = "$BackupJob_list"; //String where I need the info from$needle = "2019-08-30"; //The search value$lastPos = 0; //Starting position$positions = array();//Get lastPos and go again...while (($lastPos = strpos($string, $needle, $lastPos))!== false) {    $positions[] = $lastPos;    $lastPos = $lastPos + strlen($needle);}//Echo every value that is found in the string.foreach ($positions as $value) {    echo substr($BackupJob_list, $value, 19)."<br/>";}我想要得到的是：2019-06-30-08-00-00现在我得到2019-06-30-08-00-002019-06-30-08-30-00

Answer 1

您可以使用strpos( doc ) 将其检查为：

$str = "2019-06-30-08-00-00,2019-07-30-08-00-00,2019-08-30-08-00-00";

$needle = "2019-08-30";

$arr = explode(",", $str); // convert string to array

foreach($arr as $e) {

    if (strpos($e, $needle) === 0) echo "Found! Here: $e \n";

}