我已經(jīng)能夠使用表下方的代碼從此表中獲取多個(gè)詳細(xì)信息......我知道該表看起來(lái)很奇怪，但這是客戶端提供的可用數(shù)據(jù)表。 id | class | gender | e1 | e2 | d1 | d2 ---------------------------------------------------------------------------------------------------------------------------------------------------------------- 1 | class1 | male | ,first, first, first | ,first, first, first | , 2019-12-24,2019-12-24,2019-12-24 | ,2019-12-04,2019-11-21,2019-11-20 | 2 | class2 | female | ,second, second, second | ,second, second, second | ,2019-12-04,2019-11-21,2019-11-20 | , 2019-12-20,2019-12-20,2019-12-20 3 | class3 | male | ,first, first, first | ,third, third, third | ,2019-12-06,2019-12-13,2019-12-19 | ,2019-12-06,2019-12-13,2019-12-19 4 | class4 | male | ,third, third, third | ,third, third, third | ,2019-12-20,2019-12-20,2019-12-20 | ,2019-12-24,2019-12-24,2019-12-24 5 | class5 | female | ,second, second | ,third, third, third | ,2019-12-24,2019-12-24,2019-12-24 | ,2019-12-24,2019-12-24,2019-12-24下面是我用來(lái)獲取詳細(xì)信息的代碼：$datfrm ="00/00/0001";$datto ="31/12/2099";$gender="male";$place_map = ['first' => 1, 'second' => 0.8, 'third' => 0.4];//e1view$query = "SELECT e1 FROM eyfstb WHERE gender = '$gender'";// perform the query ...if ($result = $db->query($query)) { $finalScore1 = 0; $scores=""; $last_score=""; /* fetch associative array */ while ($row = $result->fetch_assoc()) { $efind = $row['e1']; if (substr_count($efind, ",") >'2'){ $scores = explode(',', $row['e1']); $last_score = trim($scores[count($scores)-1]); $finalScore1 += $place_map[$last_score]; } }}
5 回答
汪汪一只貓
TA貢獻(xiàn)1898條經(jīng)驗(yàn) 獲得超8個(gè)贊
您想要跳過(guò)字段中不正好有 3 個(gè)元素的行e1。在行：之后?$efind = explode(',', $efind);，添加以下內(nèi)容：
if?(count($efind)?!==?3)?{
????continue;
}在兩個(gè)地方都添加這個(gè)。
代碼中的另一個(gè)問(wèn)題是，當(dāng)您嘗試比較日期時(shí)，您將字符串與<=和進(jìn)行比較>=。您應(yīng)該使用strcmp()。
如果不知道表的全部?jī)?nèi)容，就很難看出未定義索引錯(cuò)誤來(lái)自何處。
梵蒂岡之花
TA貢獻(xiàn)1900條經(jīng)驗(yàn) 獲得超5個(gè)贊
首先，更改您的日期
$datfrm ="00/00/0001";
$datto ="31/12/2099";
到
$datfrm ="00-00-0001";
$datto ="31-12-2099";
然后加
if (empty($last_score)) {
$last_score=0;
}else{
}
之間
$last_score = trim($scores[count($scores)-1]);
和
$finalScore15 += $place_map[$last_score];
那應(yīng)該可以完全解決你的問(wèn)題。
慕桂英3389331
TA貢獻(xiàn)2036條經(jīng)驗(yàn) 獲得超8個(gè)贊
你正在做
if ($result = $db->query($query)) {
$finalScore1 = 0;
......
}
if ($result = $db->query($query)) {
$finalScore2 = 0;
.......
}
echo ($finalscore +$finalscore2);
試著做：
$finalScore1 = 0;
if ($result = $db->query($query)) {
......
}
$finalScore2 = 0;
if ($result = $db->query($query)) {
.......
}
echo ($finalscore +$finalscore2);
你還沒(méi)有定義，在所有上面定義：
$place_map = ['first' => 1, 'second' => 0.8, 'third' => 0.4];
希望這些可以解決您的錯(cuò)誤，根據(jù)數(shù)據(jù)和代碼休息結(jié)果
DIEA
TA貢獻(xiàn)1820條經(jīng)驗(yàn) 獲得超2個(gè)贊
我希望這個(gè)能幫上忙
<?php
$Date1 = '01-10-2010';
$Date2 = '07-12-2010';
$array = array();
$range = 0;
$Variable1 = strtotime($Date1);
$Variable2 = strtotime($Date2);
for ($currentDate = $Variable1; $currentDate <= $Variable2;
$currentDate += (86400)) {
$Store = date('Y-m-d', $currentDate);
$array[] = $Store;
$range++;
}
for ($i=0; $i < $range; $i++) {
echo $array[$i];
}
?>
MYYA
TA貢獻(xiàn)1868條經(jīng)驗(yàn) 獲得超4個(gè)贊
我在理解你想要從 e1 和 e2 中得到什么時(shí)遇到了一些困難 - 假設(shè)它只是該字段中的最后一個(gè)值。如果我錯(cuò)了，你可以在函數(shù) $eview 中更正它。
下面的代碼使用您發(fā)布的數(shù)據(jù)產(chǎn)生 4.2 的結(jié)果。
$datfrm = ""; // or something like '2010-01-01'
$datto = '2200'; // or something like '2015-12-01'
$gender="male";
$place_map = ['first' => 1, 'second' => 0.8, 'third' => 0.4];
// My understanding of the code is to get the weighted value of the
// last word in the column - if that's wrong, you can change this
// I am assuming you want the date that corresponded to the item you used
// for the final score, so if wrong, you need to change this.
// Also assume there will be the same number of dates as 'e's
//
// return false if bad value, or 0 if not in date range
$eview = function ($row, $ecol, $dcol) use ($place_map, $datfrm, $datto) {
$parts = array_filter(array_map('trim', explode(",", $row[$ecol])));
$dates = array_filter(array_map('trim', explode(",", $row[$dcol])));
if (!$parts || count($parts) != count($dates)) {
echo "Expected same number of parts and dates!\n";
return false;
}
$date = array_pop($dates); // Getting last date...
return ($date >= $datfrm && $date < $datto) ?
($place_map[array_pop($parts)] ?? false) : 0;
};
$finalScore1 = $finalScore2 = 0;
// perform the query ...
$query = "SELECT e1, e2, d1, d2 FROM eyfstb WHERE gender = '{$gender}'";
if ($result = $db->query($query)) {
/* fetch associative array */
while ($row = $result->fetch_assoc()) {
$finalScore1 += $eview($row, 'e1', 'd1') ?: 0;
$finalScore2 += $eview($row, 'e2', 'd2') ?: 0;
}
}
echo ($finalScore1+$finalScore2) . "\n";
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