当我从mysql显示数据时,我收到此警告:
警告:mysql_fetch_assoc()期望参数1为资源,布线在第821行的/home/zh004600/www_root/public/text/atomia/find.php中给出
但我看到了结果。如果没有这个警告我该怎么办?第821行与while循环“while($ row1 = mysql_fetch_assoc($ result1)){”
有人可以帮忙吗?
echo '<table class="files" cellspacing="3" cellpadding="3">';
$query =
'SELECT * FROM files WHERE cz ='.$Podp;
$result = mysql_query($query);
$resultCount = mysql_num_rows($result) + 3;
for ($k=0;$k<$resultCount;$k+=4) {
$j= $k-4;
echo '<tr class="files">';
if ($j == 0) {
$query1 = 'SELECT * FROM files WHERE cz ='.$Podp.' LIMIT 4';
} else {
$query1 = 'SELECT * FROM files WHERE cz ='.$Podp.' LIMIT 4 OFFSET '.$j;
}
$result1 = mysql_query($query1);
while ($row1 = mysql_fetch_assoc($result1)) {
$nazov = $row1['nazov'];
$cesta = $row1['cesta'];
echo '<th class="files"><a href="'.$cesta.'" download><img src="img/icons/pdf.png" width="45"></a><br>'.$nazov.'</th>';
}
echo '</tr>';
}
echo '</table>';
答案 0 :(得分:2)
只需添加此支票
即可if($result1){
像这样:
$result1 = mysql_query($query1);
if($result1){ // <-- THIS CHECK
while ($row1 = mysql_fetch_assoc($result1)) {
$nazov = $row1['nazov'];
$cesta = $row1['cesta'];
echo '<th class="files"><a href="'.$cesta.'" download><img src="img/icons/pdf.png" width="45"></a><br>'.$nazov.'</th>';
}
}