Question

我有两张桌子（顾客和房地产）。在客户表中，我有客户的个人数据和他的兴趣。

表“dbc_customers”：

+----+-------+-----------------+---------+------+--------+-----------+-----------+--------+
| id | name  |      email      | bedroom | bath | garage | min_price | max_price | status |
+----+-------+-----------------+---------+------+--------+-----------+-----------+--------+
|  1 | Maria | maria@email.com |       4 |    2 |      0 | 0.00      | 0.00      |      1 |
|  2 | John  | john@email.com  |       4 |    0 |      0 | 0.00      | 0.00      |      1 |
|  3 | Julia | julia@email.com |       0 |    0 |      0 | 0.00      | 0.00      |      1 |
|  4 | Ana   | ana@email.com   |       0 |    0 |      0 | 0.00      | 0.00      |      0 |
+----+-------+-----------------+---------+------+--------+-----------+-----------+--------+

在表房地产中我有登记家的每个数据。

表“dbc_posts”：

+----+------+---------+---------+------+--------+-------------+------------+--------+
| id | city | address | bedroom | bath | garage | total_price | year_built | status |
+----+------+---------+---------+------+--------+-------------+------------+--------+
|  1 |    3 | st 21   |       4 |    2 |      1 | 200.00      |       2010 |      1 |
|  2 |    3 | st 22   |       4 |    3 |      4 | 10.00       |       2000 |      1 |
|  3 |    3 | b 12    |       2 |    1 |      5 | 40.00       |       2014 |      1 |
|  4 |    2 | b 14    |       3 |    2 |      2 | 30.00       |       2013 |      1 |
+----+------+---------+---------+------+--------+-------------+------------+--------+

我需要以某种方式比较每个客户与每个家庭的利益，并显示与每个客户兼容的房屋数量，结果将是这样的：

Client1 || cliente1@email.com || 4 properties compatible
Client2 || cliente2@email.com || 7 properties compatible

然而已经尝试了各种形式，我已经打破了头脑，我已经得到了类似的结果，但总有一些错误。

在下面的代码中，它正确计算了与每个客户端兼容的房屋数量，但它也显示了兴趣空白的客户，我只需要展示满足兴趣并向其展示兼容房屋的客户。此代码可以正常显示所有客户，即使他们的利益空洞。

我目前的代码：

<?php
#Select all active customers and order by id desc
$query = mysql_query("SELECT * FROM dbc_customers WHERE status='1' ORDER BY id DESC") or die(mysql_error());
#No customers found
if (mysql_num_rows($query) < 1){
    echo "No customers found!";
}
else {
    #Set vars
    $where="";
    $i=1;
    while ($row = mysql_fetch_object($query)) {
        #Define "where" clause according to values of the table column
        if (!empty($row->bedroom)) $where .= "bedroom='$row->bedroom' AND ";
        if (!empty($row->bath)) $where .= "bath='$row->bath' AND ";
        //if (!empty($row->garage)) $where .= "c.garage = p.garage AND ";
        #Count all posts compatibles with each customer
        $query2 = mysql_query("SELECT id FROM dbc_posts WHERE $where status='1'") or die(mysql_error());
        #If none posts found break the loop, exit and show a message error, else show number of posts found
        if (mysql_num_rows($query2) < 1){ break; exit; } else { $result = mysql_num_rows($query2); }
        #Select only one post compatible for each customer
        $query3 = mysql_query("SELECT DISTINCT id FROM dbc_posts WHERE $where status='1' LIMIT 1") or die(mysql_error());
        #Flag for where var
        if ($query2 and $query3) $where = "";
        #Loop for each result of query3 and show customers and yours compatibles posts
        while ($row3 = mysql_fetch_object($query3)) {
            #Show customers
            echo "<b>".$row->name."</b> || ".$row->email." || <a href='#'><b>".mysql_num_rows($query2)." properties compatible</b></a><br />";
        }
    }
    #If none compatibles posts with customers was found
    if ($result < 1){
        echo "No listings were found compatible with any client!";
    }
}
?>

我相信我的代码可能完全错误来自以下query3变量。

Answer 1

我根本不明白为什么你需要查询3。此外，如果您在找到没有兼容性的客户后中断并退出，您将无法看到任何其他客户仍然可能具有兼容性并且未能看到任何错误消息。你需要使用＆＃39;继续＆＃39;而是去找下一位顾客。

试试这个：

#Select all active customers and order by id desc
$query = mysql_query("SELECT * FROM dbc_customers WHERE status='1' ORDER BY id DESC") or die(mysql_error());
#No customers found
if (mysql_num_rows($query) < 1) echo "No customers found!";
else {
    #Set vars
    $where = "";
    $total = 0;
    while ($row = mysql_fetch_object($query)) {
        #Define "where" clause according to values of the table column
        if ($row->bedroom > 0) $where .= "bedroom='$row->bedroom' AND ";
        if ($row->bath > 0) $where .= "bath='$row->bath' AND ";
        #Count all posts compatibles with each customer
        $query2 = mysql_query("SELECT id FROM dbc_posts WHERE $where status='1'") or die(mysql_error());
        $count = mysql_num_rows($query2);
        #If none posts found continue, else update total number of compatibles
        if(!$count) continue;
        else $total += $count;
        #Show customers
        echo "<b>".$row->name."</b> || ".$row->email." || <a href='#'><b>".$count." properties compatible</b></a><br />";
    }
    #If none compatibles posts with customers was found
    if ($total < 1) echo "No listings were found compatible with any client!";
}

顺便说一句，为什么你需要＆＃39; a＆＃39;兼容性标签？

比较2个mysql表

1 个答案: