(发布到 Stack 的第一个问题)
我正在尝试创建一个函数:
- 搜索具有变量值的字符串
- 找到该变量值后,它将删除该变量值之前的所有内容
- 它将删除以下预定字符串之后的所有内容
这是代码:
<?php
$user_id = '5';
$week_start = "2014-8-01 00:00:00";
$week_end = "2014-09-01 23:59:59";
//get the row information
$result_show_picks = mysql_query("
SELECT
id, uid, pick, date_submitted
FROM picks_recorded
WHERE uid = '" . $user_id . "'
AND date_submitted > '" . $week_start . "'
AND date_submitted < '" . $week_end . "'
LIMIT 1
", $connection);
if (!$result_show_picks) {
die("Database query failed: " . mysql_error());
}
while ($row_picks = mysql_fetch_array($result_show_picks)) {
echo "id = " . $row_picks[id] . "<br />";
echo "uid = " . $row_picks[uid] . "<br />";
echo "pick = " . $row_picks[pick] . "<br />";
echo "date_submitted = " . $row_picks[date_submitted] . "<br />";
/* this is the output
id = 233
uid = 5
pick = |1046:648|145:66|1348:736|506:334|97:37|710:434|1421:768|361:257|325:235|698:430|1457:796|1142:694
date_submitted = 2014-08-18 02:50:00
*/
$pick_input = $row_picks[pick];
};
function assign_values($pick_input) {
$str1 = ":";
$str2 = "|";
$team_picked = substr($pick_input,$str1,$str2);
return
$var1 = value1,
$var2 = value2,
$var3 = value3,
etc, etc;
/* hoped for return will be (but I have no idea if this is even feasible):
$pick_1046 = "648",
$pick_145 = "66",
$pick_1348 = "736",
etc, etc;
*/
};
?>
这就是我想要的使用方式:
<?php
if($row['this_games_id'] == $pick_1046){
echo "yes";
};
?>
我希望这是有道理的。如果重要的话,这是一个足球精选网站,仅供 friend 使用,不涉及任何形式的赌博。
感谢您的帮助。
最佳答案
对您想要从中获取信息的每个字符串执行以下操作,您将获得 %start%
和 %end%
之间的任何内容。
$start = '%start%';
$end = '%end%';
$long_string = 'Lorem ipsum dolor sit amet, consectetur adipisicing elit.%start%I want this text%end% Tempore quae explicabo tempora eum aspernatur harum, rem itaque cumque ipsum neque!';
$pos1 = (stripos($long_string, $start) + strlen($start));
$pos2 = (stripos($long_string, $end));
$result = substr($long_string, $pos1, ($pos2 - strlen($long_string)));
echo $result;
关于php - 创建一个查找函数来获取字符之间的字符串,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25359897/