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我有這樣的PHP代碼從數據庫的SQLite3 ::查詢語法錯誤
<?php
require_once ("db.php");
$db = new MyDb();
if (isset($_POST['limit']) && isset($_POST['start'])) {
$start = $_POST["start"];
$limit = $_POST["limit"];
$query =<<<EOF
SELECT * FROM questions ORDER BY quiz_id DESC '$start', '$limit';
EOF;
$result = $db->query($query);
while ($row = $result->fetchArray(SQLITE3_ASSOC)) {
echo 'div class="quesbox">
<div class="questitle">
<h2>'.$row["question"].'</h2>
</div>
<div class="quesanswer">'.$row["answer"].'</div>
<div class="quesdatetime"><img src="images/questime.png" alt="export question">'.$row["date"].'</div>
</div>';
}
}
?>
獲得的數據,但我每次運行這段代碼時我得到這些錯誤
Warning: SQLite3::query(): Unable to prepare statement: 1, near "'0'": syntax error in C:\xampp\htdocs\xport\searchfetch.php on line 14
Fatal error: Call to a member function fetchArray() on a non-object in C:\xampp\htdocs\xport\searchfetch.php on line 16
我已經嘗試了所有我知道通過編輯query
聲明來解決問題的可能方式,但無濟於事。請從哪裏來的問題。任何幫助,將不勝感激。
請做'回聲$查詢;'。顯然,查詢有問題。 –
這是'echo'結果'SELECT * FROM questions ORDER BY quiz_id DESC'0','7';'我試圖從數據庫中獲取數據的極限,其中7是極限 – diagold