作爲MySQL SELECT語句的一部分,我試圖動態地查詢數據庫。PHP implode()用「數組」替換所有數組值
當前代碼:
$stmt = $pdo->prepare('SELECT `COLUMN_NAME` FROM `INFORMATION_SCHEMA`.`COLUMNS` WHERE `TABLE_SCHEMA`="assignment" AND `TABLE_NAME`=:table AND `COLUMN_NAME` != :column1 AND `COLUMN_NAME` != :column2;');
$criteria = [
'table' => $_GET['section'],
'column1' => 'jobid',
'column2' => 'catid'
];
$stmt->execute($criteria);
$arr = array();
echo '<form method="POST">';
foreach ($stmt as $row){
echo '<label>'.ucwords($row['COLUMN_NAME']).':</label>
<input type="text" name="'.$row['COLUMN_NAME'].'"/><p>';
$arr[] = $row;
}
echo '<input type="submit" value="Submit" name="submit"/>
</form>';
if (isset($_POST['submit']))
echo $query = implode(',', $arr);
我找到了使用$ _ POST值工作得很好,但由於某種原因,它輸出:
Array,Array,Array,Array,Array
即使$改編的的var_dump()是:
0 =>
array (size=2)
'COLUMN_NAME' => string 'title' (length=5)
0 => string 'title' (length=5)
1 =>
array (size=2)
'COLUMN_NAME' => string 'salary' (length=6)
0 => string 'salary' (length=6)
2 =>
array (size=2)
'COLUMN_NAME' => string 'location' (length=8)
0 => string 'location' (length=8)
3 =>
array (size=2)
'COLUMN_NAME' => string 'description' (length=11)
0 => string 'description' (length=11)
4 =>
array (size=2)
'COLUMN_NAME' => string 'category' (length=8)
0 => string 'category' (length=8)
implode()在多維數組上不起作用。 – developerwjk
您的var_dump顯示$ arr是數組的數組。那是你要的嗎? –
這是僞代碼,因爲我可以看到它不應該超過第一個查詢的一些原因? – RiggsFolly