1

MySQL中有三个表：Employees，Branches，Departments。我需要信息以下列方式出现：

亚特兰大分公司
交付部
Phillip J. Fry
电话：123456

工程部
Turanga Leela
电话：123457

Bender Rodriguez
电话：123458


当前简单的 PHP 代码：
1）从三个表中获取行（使用 JOIN 进行简单的 SELECT 查询）
2）将它们放在行中（mysql_fetch_assoc）
3）使用 PHP While 循环

显示结果是这样的：

亚特兰大分公司
交付部
Phillip J. Fry
电话：123456

亚特兰大分公司
工程部
Turanga Leela
电话：123457

亚特兰大分公司
工程部
Bender Rodriguez
电话：123458

你可以推荐什么技术（JS、jQuery、Ajax）或方法，这样我就可以只使用一个查询来提取行信息，而不是重复分支名称和部门名称？

更新：如果我将分支名称放在循环之外（使用 While 循环），则会有多个循环：1）获取分支，2）获取部门，3）获取该部门的所有员工。环形。

更新：共享代码： 

<?php
// Create connection
$connection = mysql_connect('localhost','root', '') or die('Connection error.');
mysql_query("SET NAMES 'utf8'", $connection);
mysql_select_db("eReference");

// Check Employees
$query = "SELECT Employees.fName, Employees.lName, Department.deptName, Branch.branchName, ".
"FROM Employees ".
"LEFT JOIN Department ".
"ON Employees.department = Department.id ".
"LEFT JOIN Branch ".
"ON Employees.branch = Branch.id ;";

$result = mysql_query($query, $connection) or die(mysql_error());

while ($row = mysql_fetch_assoc($result)) {

?>

<h2><?php echo $row['branchName']; ?></h2>
<?php if ($row['deptName']) echo "<h3>" . $row['deptName'] . "</h3>"; ?>
<h4><?php echo $row['fName'] . " " . $row['lName']; ?></h4></p>
<?php
}
?>
4

6 回答 6

2 
<?php

   $i = 1; // to be incremented after printing branchName once

   while ($row = mysql_fetch_assoc($result)) { 

     if($i == 1) { ?>

        <h2><?php echo $row['branchName']; $i ++;  ?></h2>

     <?php } ?>

     <?php if ($row['deptName']) echo "<h3>" . $row['deptName'] . "</h3>"; ?>

     <h4><?php echo $row['fName'] . " " . $row['lName']; ?></h4></p>

   <?php } ?>

只需添加一个变量$i = 1并在打印前检查它是否等于 1。第一次打印后，将其递增。它只是添加了一个if语句。

希望这可以帮助。

于 2013-08-06T11:07:20.353 回答
1

我就是这样做的。

使用数据创建一个多维数组，并遍历该数组以呈现输出。

这在内存使用方面并不是最有效的，但除非您有数千行数据，否则它可能不会成为问题。

这样做的好处，就是html渲染代码更加简单易懂，而且sql和html没有混在一起。（有利于代码维护）

<?php
// Create connection
$connection = mysql_connect('localhost','root', '') or die('Connection error.');
mysql_query("SET NAMES 'utf8'", $connection);
mysql_select_db("eReference");

// Check Employees
$query = "SELECT Employees.fName, Employees.lName, Department.deptName, Branch.branchName, ".
    "FROM Employees ".
    "LEFT JOIN Department ".
    "ON Employees.department = Department.id ".
    "LEFT JOIN Branch ".
    "ON Employees.branch = Branch.id ;";
// note you probably want to add an order by statement here too, to ensure consistent sorting

$result = mysql_query($query, $connection) or die(mysql_error());

$data = array();

// build a multi-dimensional array from the result set
while ($row = mysql_fetch_assoc($result)) {

    $data[($row['branchName'])][($row['deptName'])][] = array(
        'name' => "{$row['fName']} {$row['lName']}",
        'phone' => $row['phone'] // add phone, doesn't exist in original query, but just to illustrate how it would work
    );
}

// sql finishes here

?>

<?php
    // html rendering 
    // use htmlentities to escape any html chars, such as < > etc

    foreach ($data as $branchName => $departments) {

        echo '<h2>',htmlentities($branchName),'</h2>';

        foreach ($departments as $deptName => $employees) {

            foreach ($employees as $employee) {
                echo '<h3>',htmlentities($deptName),'</h3>';
                echo '<h4>',htmlentities($employee['name']),'</h4>';
                echo '<h4>',htmlentities($employee['phone']),'</h4>';
            }
        }
    }
?>
于 2013-08-07T04:09:37.130 回答
0

也许这会起作用..

<?php
$deptName;
while($row = mysql_fetch_assoc($result))
{
    if ($row['deptName'])
    {
        if ($deptName != $row['deptName'])
        {
            echo "<h3>" . $row['deptName'] . "</h3>";
                        $deptName = $row['deptName'];
        }
    }
}
?>
于 2013-08-06T11:16:54.283 回答
0

从你的 sql 得到的结果应该是一个数组，所以使用 while 循环迭代抛出数组，同时回显当前索引的结果

于 2013-08-06T10:54:40.513 回答
0 
   $result = mysql_query($query, $connection) or die(mysql_error());
  $newarray = array();
  $i = 0;
while ($row = mysql_fetch_assoc($result)) {
 $newarray[$i]['deptName'] = $row['deptName'];
 $newarray[$i]['fName'] = $row['fName'];
$newarray[$i]['lName'] = $row['lName'];
$i++; 
}
<h2><?php echo $newarray[0]['branchName']; ?></h2>
while($newarray){
?>
<?php if ($newarray['deptName']) echo "<h3>" . $newarray['deptName'] . "</h3>"; ?>
<h4><?php echo $newarray['fName'] . " " . $newarray['lName']; ?></h4></p>
}

?>

于 2013-08-06T11:00:08.957 回答
0

所以，这就是它所做的：

$query = "SELECT Employees.lName, Employees.fName, Employees.mName, Position.position, ".
                    "department.deptName, department.deptId, ".
                    "Branch.branchName, Branch.branchId, ContactInformation.* ".
                    "FROM Employees ".
                    "LEFT JOIN Position ".
                    "ON Employees.position = Position.id ".
                    "LEFT JOIN Department ".
                    "ON Employees.department = Department.deptId ".
                    "LEFT JOIN Branch ".
                    "ON Employees.branch = Branch.branchId ".
                    "LEFT JOIN ContactInformation ".
                    "ON Employees.contactInformation = ContactInformation.id ".
                    "ORDER BY Employees.branch, Employees.department ASC;";

            $result = mysql_query($query, $connection) or die(mysql_error());

            $arrayOfEmployees = array();

            while ($row = mysql_fetch_assoc($result)) {
                $arrayOfEmployees[($row['branchName'])][($row['deptName'])][] = array(
                    'lName' => $row['lName'],
                    'fName' => $row['fName'],
                    'mName' => $row['mName'],
                    'position' => $row['position'],
                    'lPhone1' => $row['lPhone1'],
                    'lPhone2' => $row['lPhone2'],
                    'mPhone' => $row['mPhone'],
                    'fax' => $row['fax'],
                    'office' => $row['office'],
                    'email' => $row['email']
                );
            }

            foreach($arrayOfEmployees as $branchName => $arrayOfDepartments) {
                echo "<h2>".$branchName."</h2>";
                foreach($arrayOfDepartments as $deptName => $arrayOfEmployeeContacts) {
                    echo '<h3>',htmlentities($deptName),'</h3>';
                    foreach($arrayOfEmployeeContacts as $employeeContacts) {
                        echo "<h4>".$employeeContacts["lName"]." ".$employeeContacts["fName"]." ".$employeeContacts["mName"]."</h4>";
                        echo "<p>";
                        if($employeeContacts["position"]) echo $employeeContacts["position"]."<br>";
                        $num = $employeeContacts["lPhone1"];
                        if($employeeContacts["lPhone1"]) echo "+".substr($num,0,1)." (".substr($num,1,4).") "." ".substr($num,5,2)."-".substr($num,7,2)."-".substr($num,9,2)."<br>";
                        $num = $employeeContacts["lPhone2"];
                        if($employeeContacts["lPhone2"]) echo "+".substr($num,0,1)." (".substr($num,1,4).") "." ".substr($num,5,2)."-".substr($num,7,2)."-".substr($num,9,2)."<br>";
                        $num = $employeeContacts["mPhone"];
                        if($employeeContacts["mPhone"]) echo "+".substr($num,0,1)." (".substr($num,1,3).") "." ".substr($num,4,3)."-".substr($num,7,2)."-".substr($num,9,2)."<br>";
                        $num = $employeeContacts["fax"];
                        if($employeeContacts["fax"]) echo "+".substr($num,0,1)." (".substr($num,1,4).") "." ".substr($num,5,2)."-".substr($num,7,2)."-".substr($num,9,2)."<br>";
                        if($employeeContacts["email"]) echo "<a href=\"mailto:".$employeeContacts["email"]."\">".$employeeContacts["email"]."</a><br>";
                        if($employeeContacts["office"]) echo "Кабинет — ".$employeeContacts["office"];
                        echo "</p>";
                    }
                }
            }

经测试。该解决方案动态运行良好——只需将更多分支机构和部门添加到数据库中。这是针对我最初使用的方法的效率检查（以微秒为单位）：

        Array-based Original
1       1.015       1.012
2       1.016       1.02
3       1.026       1.013
4       1.015       1.002
5       1.026       1.02
6       1.014       1.02
7       1.013       1.019
8       1.005       1.014
9       1.013       1.006
10      1.021       1.015
Average 1.0164      1.0141
于 2013-08-07T03:20:18.087 回答