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「米收到這個警告：mysqli_query（）：無法在/home/u443228523/public_html/easy.php取的mysqli第6行警告：mysqli_fetch_array（）
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in /home/u443228523/public_html/easy.php on line 12
從這個代碼：
<?php
// perform the query and store the result
$result = mysqli_query($conn,"SELECT * FROM library, crime_data WHERE crime_data.id=$oreo AND crime_data.isbn=library.isbn AND crime_data.visibility='0'");
echo "<iframe src='demo.html' name='fool_iframe' id='fool_iframe' style='visibility:hidden; display:none;'></iframe>
<form method='post' action='delete.php' name='erradica'>";
echo "<ol id='printy'>";
while($row = mysqli_fetch_array($result))
{
echo "<li class='daitems'>
<div class='label'>
<a href='".$row['url']."' target='_blank'><img src='". $row['src'] . "' width='20' height='30' alt='".$row['title']."' style='float:left'></a>
<a href='delete.php?id_line2=".$row['url']."' onClick='JavaScript:timedRefresh(500);' target='fool_iframe' style='float:right;'><img src='images/close_window.png'></a>
<a href='".$row['url']."' target='_blank' class='sec_par'><span id='title33'>".$row['title']." </span>
<p style='margin:0; padding:0;'>
".$row['url']." <br/> </a>
</p>
</div>
</li>";
}
echo "</ol>
</form>";
?>
有人可以解釋我爲什麼嗎？它可以在數據庫設置上嗎？
BTW $ oreo在conexion.php上聲明。
'echo「SELECT * FROM library，crime_data WHERE crime_data.id = $ oreo AND crime_data.isbn = library.isbn AND crime_data.visibility ='0'」'檢查查詢是否正在輸出您認爲應該是（和張貼在這裏）也使用'mysqli_error（$ conn）'在查詢後查看任何錯誤（也張貼在這裏） – Steven
感謝您的幫助：）不知道如何，但現在工作，也許只是服務器緩存 –