Question

所以我之前写过（在php中），但每次我尝试echo $ test“，我只是回到资源id 5.有谁知道如何从变量中实际打印出mysql查询？

$dave= mysql_query("SELECT order_date, no_of_items, shipping_charge, SUM(total_order_amount) as test FROM `orders` WHERE DATE(`order_date`) = DATE(NOW()) GROUP BY DATE(`order_date`)") or die(mysql_error());
print $dave;

Answer 1

这将打印出查询：

$query = "SELECT order_date, no_of_items, shipping_charge, SUM(total_order_amount) as test FROM `orders` WHERE DATE(`order_date`) = DATE(NOW()) GROUP BY DATE(`order_date`)";

$dave= mysql_query($query) or die(mysql_error());
print $query;

这将打印出结果：

$query = "SELECT order_date, no_of_items, shipping_charge, SUM(total_order_amount) as test FROM `orders` WHERE DATE(`order_date`) = DATE(NOW()) GROUP BY DATE(`order_date`)";

$dave= mysql_query($query) or die(mysql_error());

while($row = mysql_fetch_assoc($dave)){
    foreach($row as $cname => $cvalue){
        print "$cname: $cvalue\t";
    }
    print "\r\n";
}

Answer 2

你正在从数据库中返回一系列项目。所以你需要这样的东西。

   $dave= mysql_query("SELECT order_date, no_of_items, shipping_charge, 
    SUM(total_order_amount) as test FROM `orders` 
    WHERE DATE(`order_date`) = DATE(NOW()) GROUP BY DATE(`order_date`)") 
    or  die(mysql_error());


while ($row = mysql_fetch_assoc($dave)) {
echo $row['order_date'];
echo $row['no_of_items'];
echo $row['shipping_charge'];
echo $row['test '];
}

Answer 3

来自php docs：

对于SELECT，SHOW，DESCRIBE，EXPLAIN和其他语句返回   resultset，mysql_query（）在成功时返回资源，或者返回FALSE   错误。

对于其他类型的SQL语句，INSERT，UPDATE，DELETE，DROP等，   mysql_query（）成功时返回TRUE，错误时返回FALSE。

返回的结果资源应该传递给mysql_fetch_array（），   以及用于处理结果表的其他函数，以访问   返回数据。

http://php.net/manual/en/function.mysql-query.php

Answer 4

$sql = "SELECT * FROM table_name ORDER BY ID DESC LIMIT 1";
$records = mysql_query($sql);

你可以改变限制1 至限制你想要的任何数字

这将首先显示最后一个INSERTED行。

从变量打印mysql查询的结果

4 个答案: