我遇到了一些我已经工作了几天的特定PHP代码的问题。这是一个报告代码,我可以输入一天一个月,它将列出该特定日期的总销售额。
但是,我似乎无法做出最后的陈述,如果查询中没有值(没有数据),它将显示“此特定日期没有销售额”。这是我一直在努力的代码。但最后一个echo语句没有执行。有什么想法吗?
<?php
session_start();
if ((isset($_SESSION["admin"])) ){
$day=@$_POST['day'];
$month=@$_POST['month'];
echo "<center><h2>Sales report on " .$day. "." .$month. ".2013</h2></center>";
echo "<center><table style='border:2px solid black;' align=center width=600>";
echo "<tr><th colspan=12><center><h2>Sales Report</h2><hr size='2' color='black' /></center></th></tr>";
echo " <th width=400> Amount Collected</th>";
?>
<br>
<?php
$x = 1; //counter
//open a connection to a MySQL server using function mysql_connect
//returns a MySQL link identifier on success, or FALSE on failure.
$conn= mysql_connect("localhost","root","");
if (!$conn)
die ("Connection error: ".mysql_error());
else {
//select a MySQL database
//returns TRUE on success or FALSE on failurue.
$db=mysql_select_db("cqfos");
if(!$db)
die ("DB not found: ".mysql_error());
else {
//put query in a variable $query
$query= "select ROUND(sum(orderdetails.tprice),2)
from orders JOIN orderdetails ON orders.orderID = orderdetails.orderID WHERE DAY(orders.date) = '$day' AND MONTH(orders.date) = '$month'";
$result=mysql_query($query);
if(!$result)
die ("Invalid query: ".mysql_error());
//if record exists
else {
//fetch a result row as both associative array and numeric array
if(mysql_num_rows($result)== 1){
while ($row=mysql_fetch_array($result,MYSQL_BOTH)){
echo "<tr>";
echo "<td align='center'>RM ".$row[0]."</td></tr>";
$x++; //increase the counter
}
}
else {
echo "<tr><th colspan=12>No sales made.</td></tr>";}
}
}
}
echo"</table></center>";
?>
答案 0 :(得分:0)
您的SQL(可能)只返回一行以上,因此请将前面提到的行更改为:
if(mysql_num_rows($result)>0){
答案 1 :(得分:0)
这里有几个问题 您的HTML表语法不正确,并且您使用的是旧的SQL库 - 它看起来不像您的SQL语法是正确的...尝试此代码(未经测试,因为我没有您的数据)
<?php
session_start();
if ((isset($_SESSION["admin"])) ){
echo '<div style="margin:auto; textalign:center;">';
echo "<h2>Sales report on " .$_POST['day']. "." .$_POST['month']. ".2013</h2>";
echo "<h2>Sales Report</h2>"
echo "<table style='border:2px solid black;' align=center width=600>";
echo "<tr><th width=400> Amount Collected</th></tr>";
?>
<br>
<?php
$conn = new mysqli("localhost","root","","cqfos");///use mysqli, not mysql : mysql is depricated
if ($conn->mysqli)
exit ("Connection error: ".$conn->errno ." : " $conn->error);
else {
//put query in a variable $query
$eDay = $conn->mysql_real_escape_string($_POST['day']);//escape these to protect the database
$eMonth = $conn->mysql_real_escape_string($_POST['month']);;//escape these to protect the database
//your column name is probably not a rounded value, replaced it with * (return all columns)
$query= "select * from orders JOIN orderdetails ON orders.orderID = orderdetails.orderID WHERE DAY(orders.date) = '"
.$eDay."' AND MONTH(orders.date) = '".$eMonth."'";
$result=$con->query($query);
if($conn->errno)
exit ("Invalid query: ".$conn->errno ." : " $conn->error);
//if record exists
else {
$numericArray = $result->fetch_array(MYSQLI_NUM); //fetch a result row as numeric array
$associativeArray = $result->fetch_array(MYSQLI_ASSOC); //fetch as an associtive array this is not used, just an example
$bothArray = $result->fetch_array(MYSQL_BOTH); //both associtive and numeric this is not used, just an example
}
if(!empty($numericArray))
{
foreach ($numericArray as $value) {
echo "<tr><td>RM ".$value[0]."</td><tr>";//is there more then 1 col? if not you should consider an html list
}
} else {
echo "<tr><td>No sales made</td><tr>";
}
echo"</table></center>";
}
?>
答案 2 :(得分:0)
只是让你知道你的代码容易受到SQLi攻击,因为你没有清理$ day和$ month。还please consider using PDO。
如果您还没有 - 尝试将SQL语句运行到PHPMyAdmin并查看它输出错误的位置(如果有),否则它将输出数据。*
*手动输入日/月代替变量。