Php SQL查询错误查询为空
您调用了两次Php SQL查询错误查询为空,php,mysql,Php,Mysql,您调用了两次mysql\u查询函数,但没有任何结果。尝试将您的线路替换为: <?php include "head-main.php"; include "conn.php"; if(isset($_GET)){ $var = $_GET["cat"]; } $data = mysql_query('SELECT kampword FROM data_dict where category=$var'); $retval = mysql_query
mysql\u查询
函数,但没有任何结果。尝试将您的线路替换为:
<?php
include "head-main.php";
include "conn.php";
if(isset($_GET)){ $var = $_GET["cat"]; }
$data = mysql_query('SELECT kampword FROM data_dict where category=$var');
$retval = mysql_query( $data, $conn );
if(! $retval ) {
die('Could not get data: ' . mysql_error());
}
while($row = mysql_fetch_assoc($retval)) {
echo "World :{$row['kampword']} <br> ".
"--------------------------------<br>";
}
?>
或
试一试
供参考。它们不再得到维护。看到了吗?相反,请了解,并使用or-将帮助您决定哪一个最适合您。您的脚本可能会有风险,您甚至可以查看使用时发生的情况。$var
是字符串吗?它有价值吗?变量不会用简单的引号来解释。可能$data=mysql\u query(“从data\u dict中选择kampword,其中category=“.$var.”);错误已经消失,但仍然无法获取数据库中的数据
<?php
if(isset($_GET["cat"])) {
$var = $_GET["cat"];
$retval = mysql_query('SELECT kampword FROM data_dict where category=$var');
}else {
$retval = false;
}
$data = mysql_query("SELECT kampword FROM data_dict where category=$var");
$data = mysql_query('SELECT kampword FROM data_dict where category="'.$var.'"');
if(isset($_GET["cat"])){ $var = $_GET["cat"]; }