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我有这样的PHP代码从数据库的SQLite3 ::查询语法错误
<?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>';
}
}
?>
获得的数据,但我每次运行这段代码时我得到这些错误
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
我已经尝试了所有我知道通过编辑query
声明来解决问题的可能方式,但无济于事。请从哪里来的问题。任何帮助,将不胜感激。
请做'回声$查询;'。显然,查询有问题。 –
这是'echo'结果'SELECT * FROM questions ORDER BY quiz_id DESC'0','7';'我试图从数据库中获取数据的极限,其中7是极限 – diagold