在數組中存儲數據庫內容PHP

Question

我不知道如何做到這一點，但如果可以做到這一點，任何人都可以幫助我。 基本上我有我的表3列：

ID Set Result Case 
1 Set1 PASS 101 
2 Set2 FAIL 102 
3 Set2 FAIL 101 
4 Set1 FAIL 101 
5 Set1 PASS 104 

$set = $row['Set']; 

我所試圖實現的是，這些Set的的foreach，存儲其關聯數組中的Result和Case。

Answer 1

$arr = array(); 

foreach ($records as $key => $value) 
{ 
    $arr[$key]['Result'] = $value['Result']; 
    $arr[$key]['Case'] = $value['Case']; 
} 

echo '<pre>'; 
print_r($arr); 
echo '</pre>'; 

在您的評論的光：

foreach ($records as $key => $value) 
{ 
    $arr[$key][$value['Result']] = $value['Case']; 
} 

在最近的評論的光：

foreach ($records as $key => $value) 
{ 
    $arr[$value['Set']][$value['Result']] = $value['Case']; 
} 

Answer 2

閱讀評論後，我想我會採取裂縫它。

首先：除非您要檢查$array[Set2][{result}]中的重複{result}鍵，並且在評論中存在重複的情況下將它們小寫，否則您要求的內容不起作用。我不知道爲什麼「 d做。這會令人困惑，並使我感到無聊。即：

$arr[Set2][FAIL]與$arr[Set2][fail]

如果做到這一點，如圖上述[在阿利克斯Axel的第三代碼塊]，你會做：

$arr[Set2][FAIL] = 102然後覆蓋該數組索引值與$arr[Set2][FAIL] = 101，導致你丟失數據。換句話說，您將「set」和「result」組合成一個「組合鍵」，可以這麼說，您不能這樣做，因爲這些組合不是唯一的（Set2 FAIL，Set2 FAIL ）。我知道這是一個令人討厭的答案，但你應該看看你在做什麼，爲什麼，因爲我有一種預感，你會以錯誤的方式去做。你可能要像一個數組：

Array 
( 
    [Set1] => Array 
    ( 
      [101] => 'FAIL' 
      [102] => 'PASS' 
    ) 

    [Set2] => Array 
    ( 
      [101] => 'FAIL' 
      [102] => 'FAIL' 
    ) 
) 

什麼的，但即使如此，當你有一些設置/案例對兩者通過和失敗，它不會工作。正因爲如此，你可以在這裏做的唯一事情就是使用「ID」作爲索引：

Array 
( 
    [1] => Array 
    ( 
      [Set] => 'Set1' 
      [Result] => 'PASS' 
      [Case] => '101' 
    ) 
    [2] => Array 
    ( 
      [Set] => 'Set1' 
      [Result] => 'FAIL' 
      [Case] => '101' 
    ) 
) 

但我甚至不能告訴你該怎麼做，因爲你還沒有告訴我們如何你查詢結果數組是在第一個地方結構！所以第1步）請print_r或var_dump查詢結果。

Answer 3

// assuming 
$myArray = array(); 
$result = mysql_query("SELECT * FROM table ORDER BY Set ASC"); 
while ($rows = mysql_fetch_assoc($result)) { 
    for ($i = 0; $i <= count($rows); $i++) { 
    $myArray[$rows['ID']]['Set'] = $rows['Set']; 
    $myArray[$rows['ID']]['Result'] = $rows['Result']; 
    $myArray[$rows['ID']]['Case'] = $rows['Case']; 
} 

// output 
Array 
(
    [1] => Array 
    (
     [Set] => 'Set1' 
     [Result] => 'PASS' 
     [Case] => '101' 
    ) 
    [2] => Array 
    (
     [Set] => 'Set1' 
     [Result] => 'FAIL' 
     [Case] => '101' 
    ) 
    [3] => Array 
    (
     [Set] => 'Set1' 
     [Result] => 'PASS' 
     [Case] => '104' 
    ) 
    [4] => Array 
    (
     [Set] => 'Set2' 
     [Result] => 'FAIL' 
     [Case] => '102' 
    ) 
    [5] => Array 
    (
     [Set] => 'Set2' 
     [Result] => 'FAIL' 
     [Case] => '101' 
    ) 
) 

Answer 4

年前寫了一個函數來處理返回MySQLs查詢結果的多維數組（名字與tablename相同）的短查詢。希望這會節省一些時間處理簡單和短期的MySQL查詢：

<?php 
/* 
FUNCTION: db2array($tablename, $condition) 

Which simply makes a query: SELECT * FROM $tablename WHERE $condition 
and returns multidimensional array()* named as 
DB_table_name with keys named as tables_cell_name followed by row_number: 

*Returns multidimensional array: $my_table[cell_name][row_number] 

Example 1: 
db2array("my_table", "cell_name = 'A' ORDER BY id LIMIT 1") 
where "my_table" is DB tablename and $condition is all the rest 
what comes after MySQL's WHERE; 

Example 2: 
db2array("my_table"); 

Returns: $my_table[]; 
To see results, go for print_r($my_table[]); 

*/ 

function db2array($tablename, $condition){ 

    global $$tablename; 

    //if $condition is NULL set it to 1=1 
    $condition = ($condition)?$condition:"1=1"; 

    $query = mysql_query(" 
    SELECT * FROM $tablename WHERE $condition 
    ") or die(mysql_error()); 

    // COUNTING FIELDS AND ROWS 
    $f_to = mysql_num_fields($query); 
    $n_to = mysql_num_rows($query); 

    // EXIT IF NO RESULTS 
    if (!mysql_num_rows($query)) return false; 


    // PUSHING ALL CELL NAMES TO $cellname ARRAY 
    for ($f=0; $f<=$f_to-1; $f++){ 
     $cellname[$f] = mysql_field_name($query, $f); 
    } 

    // GENERATING OUTPUT ARRAY 
    for ($n=0; $n<=$n_to-1; $n++){ 
     foreach ($cellname as $val){ 
      $result[$val][$n] = mysql_result($query, $n, $val); 
     } 
    } 

    $$tablename = $result; 
    return $$tablename; 

} 
?>