[英]Multiple database row values in single variable
我将所有名称添加到单个变量中,但它只显示一个值,最后一个值。
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;
不,您只是分配变量,请使用它加上一个“”。 装备前
$csk .= "'".$row['NAME']."',";
但是我建议使用数组,这样您就可以将其用于JS(if ajax)或php进行更灵活的处理
$csk = array();
while ($row = mysql_fetch_assoc($query)) {
$csk[] = array($row['NAME']);
}
echo $csk; //for ajax use echo json_encode($csk);
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 $csk;
您将在每次循环迭代时将变量重置为$row['NAME']
的值。
您需要做的是将变量附加到$csk
:
$csk .= "'".$row['NAME']."',";
^---- notice the extra . here
额外的.
表示您要将值附加到$csk
。
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