Question

我试图用PHP＆amp;我无法理解的MySQL。

我有一个采用以下结构的表（它们是两个表，但在SQL查询中加入时将作为一个表）：

id | brand  | model  
1  | brand1 | model1  
2  | brand1 | model2  
3  | brand1 | model3  
4  | brand2 | modelA

我已经能够让PHP输出以下内容：

brand1 model1  
brand1 model2  
brand1 model3  
brand2 modelA

我真正希望能够做的是将其输出为：

Brand1  
Model1  
Model2  
Model3  

Brand2  
ModelA

有人有任何想法吗？

这是我目前的代码：

<?php 
$result = mysql_query("SELECT cameras.cameraid, cameras.model, brands.brand FROM cameras JOIN brands ON cameras.brandid=brands.brandid WHERE cameras.categoryid='$cat' ORDER BY brands.brand ASC, cameras.level ASC")or die(mysql_error());

while($row = mysql_fetch_array($result))
{
     echo "<li><a href='camera.php?id=". $row['cameraid'] . "'>" . $row['brand'] . " " . $row['model'] . "</a></li>";
}

?>

Answer 1

您正在寻找的是具有嵌套foreach - 循环的二维数组。这将允许您拥有一个看起来像这样的数组

Array ( 
    [brand1] => Array ( 
        [1] => model1 
        [2] => model2 
        [3] => model3 
    ) 
    [brand2] => Array ( 
        [4] => modelA 
        [5] => modelB 
        [6] => modelC 
    ) 
)

当我们稍后将模型作为每个品牌的列表回应时，我们将此用于我们的优势。第一个数组将包含所有品牌，而每个品牌都是自己的数组 - 创建二维数组，其中包含所有模型（以及产品ID作为该数组中元素的键）。

将每个元素放入while($row = mysql_fetch_array($result))循环中的数组中，并使用嵌套的foreach循环来回显它们。

while ($row = mysql_fetch_array($result)) {
    // Here we fetch all the results, and put it into a two-dimentional array 
    // with keys as brandname and the ID of the product
    $brands[$row['brand']][$row['cameraid']] = $row['model'];
}

// Then we echo them all out in a list!
foreach ($brands as $key=>$brand) {
    // The $key is the name of the brand
    echo $key."<ul>";
    foreach ($brand as $id=>$model) {
        // The $id is the ID of the camera in the table
        // The $model is the model-name itself
        echo "<li><a href='camera.php?id=".$id."'>".$model."</a></li>";
    }
    echo "</ul>";
}

此外，为了使用mysql_*功能，已弃用且不推荐使用mysqli_*功能，您应考虑转到PDO或<?php $server = “SQLINSTANCE”; $options = array( "UID" => “uname”, "PWD" => “password, "Database" => "HomeSensors"); $conn = sqlsrv_connect($server, $options); if ($conn === false) die("<pre>".print_r(sqlsrv_errors(), true)); echo "Successfully connected!<br/>"; $stmt = mssql_init('addSensorData', $conn); mssql_bind($stmt, '@Pool', 90, SQLINT1, false, false, 3); mssql_bind($stmt, '@Temp', 80, SQLINT1, false, false, 3); $proc_result = mssql_execute($stmt); if( $proc_result === false) { die( print_r( sqlsrv_errors(), true) ); } mssql_free_statement($stmt); ?>以获得更好的功能，持续支持和更好的安全性。

Answer 2

你应该尝试这样的事情。它使用多维数组对项目进行分组。

<?php 
$result = mysql_query("SELECT cameras.cameraid, cameras.model, brands.brand FROM cameras JOIN brands ON cameras.brandid=brands.brandid WHERE cameras.categoryid='$cat' ORDER BY brands.brand ASC, cameras.level ASC")or die(mysql_error());

$brands = array();
while($row = mysql_fetch_array($result))
{
    $brands[$row['brand']][] = array($row['model'], $row['cameraid']);
    echo "<li><a href='camera.php?id=". $row['cameraid'] . "'>" . $row['brand'] . " " . $row['model'] . "</a></li>";
}

foreach($brands as $brand => $data){
    echo $brand . "<br>";
    foreach($data as $item){
        echo "<li><a href='camera.php?id=" . $item[1] . "'>" . $item[0] . "</a></li>";
    }
}

?>

Answer 3

下面是网格使用表的代码它将为您提供并了解如何实现您的任务

$sql = mysql_query("SELECT id, member_name FROM member_table ORDER BY id DESC LIMIT 15"); 
$i = 0;
// Establish the output variable
$dyn_table = '<table border="1" cellpadding="10">';
while($row = mysql_fetch_array($sql)){ 

    $id = $row["id"];
    $member_name = $row["member_name"];

    if ($i % 3 == 0) { // if $i is divisible by our target number (in this case "3")
        $dyn_table .= '<tr><td>' . $member_name . '</td>';
    } else {
        $dyn_table .= '<td>' . $member_name . '</td>';
    }
    $i++;
}
$dyn_table .= '</tr></table>';

此代码的逐行说明位于以下链接

数组网格输出教程会给你一些想法尝试将此用于你赢得的scnario Array Grid output tutorial well explained video

PHP MySQL将输出拆分为多个部分

3 个答案: