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“米收到这个警告:mysqli_query():无法在/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
从这个代码:
<?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>";
?>
有人可以解释我为什么吗?它可以在数据库设置上吗?
BTW $ oreo在conexion.php上声明。
'echo“SELECT * FROM library,crime_data WHERE crime_data.id = $ oreo AND crime_data.isbn = library.isbn AND crime_data.visibility ='0'”'检查查询是否正在输出您认为应该是(和张贴在这里)也使用'mysqli_error($ conn)'在查询后查看任何错误(也张贴在这里) – Steven
感谢您的帮助:)不知道如何,但现在工作,也许只是服务器缓存 –