php - 单个变量中的多个数据库行值

Question

我将所有名称添加到单个变量中，但最后一个仅显示一个值。

我的代码是：

include 'dbconnect.php';
$result = mysql_query("SELECT * FROM bookedtates WHERE SID='$ServiceHosterIdv' AND BOOKEDDATE='$q'");
//$result = mysql_query("SELECT * FROM bookedtates WHERE SID='$ServiceHosterIdv' AND BOOKEDDATE='$q'");
while ($row = mysql_fetch_assoc($query)) {
$csk = "'".$row['NAME']."',";
}
echo $csk; 

Answer 1

不，你只是分配变量使用它来加上一个“。” 等式之前

  $csk .= "'".$row['NAME']."',";

但我建议使用数组，这样你就可以使用 JS（如果 ajax）或 php 来获得更灵活的东西

$csk = array();
while ($row = mysql_fetch_assoc($query)) {
$csk[] = array($row['NAME']);
}
echo $csk; //for ajax use echo json_encode($csk);

Answer 2

只是测试

include 'dbconnect.php';
$result = mysql_query("SELECT * FROM bookedtates WHERE SID='$ServiceHosterIdv' AND BOOKEDDATE='$q'");
//$result = mysql_query("SELECT * FROM bookedtates WHERE SID='$ServiceHosterIdv' AND BOOKEDDATE='$q'");
$csk = '';
while ($row = mysql_fetch_assoc($query)) {
    $csk .= "'".$row['NAME']."',";
}
echo $csk; 

Answer 3

您正在将变量重置为$row['NAME']循环每次迭代的值。

您需要做的是将变量附加到末尾$csk：

$csk .= "'".$row['NAME']."',";
     ^---- notice the extra . here

额外的.表示您要将值附加到$csk.