php 从变量打印mysql查询的结果
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Printing result of mysql query from variable
提问by Louis Stephens
So I wrote this earlier (in php), but everytime I try echo $test", I just get back resource id 5. Does anyone know how to actually print out the mysql query from the variable?
所以我早些时候（在 php 中）写了这个，但是每次我尝试 echo $test" 时，我都会返回资源 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;
回答by Jeremy Holovacs
This will print out the query:
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;
This will print out the results:
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";
}
回答by Lucas
well you are returning an array of items from the database. so you need something like this.
好吧，您正在从数据库中返回一组项目。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 '];
}
回答by smp7d
From php docs:
来自 php 文档：
For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error.
For other type of SQL statements, INSERT, UPDATE, DELETE, DROP, etc, mysql_query() returns TRUE on success or FALSE on error.
The returned result resource should be passed to mysql_fetch_array(), and other functions for dealing with result tables, to access the returned data.
对于 SELECT、SHOW、DESCRIBE、EXPLAIN 和其他返回结果集的语句，mysql_query() 在成功时返回资源，在错误时返回 FALSE。
对于其他类型的 SQL 语句，INSERT、UPDATE、DELETE、DROP 等，mysql_query() 在成功时返回 TRUE，在错误时返回 FALSE。
返回的结果资源应该传递给mysql_fetch_array()和其他处理结果表的函数，以访问返回的数据。
回答by Nazmul Hasan
$sql = "SELECT * FROM table_name ORDER BY ID DESC LIMIT 1";
$records = mysql_query($sql);
you can change LIMIT 1 to LIMIT any number you want
您可以将 LIMIT 1 更改为 LIMIT 任何您想要的数字
This will show you the last INSERTED row first.
这将首先显示最后一个 INSERTED 行。
