[英]Multiple database row values in single variable
我將所有名稱添加到單個變量中,但它只顯示一個值,最后一個值。
我的代碼是:
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,您只是分配變量,請使用它加上一個“”。 裝備前
$csk .= "'".$row['NAME']."',";
但是我建議使用數組,這樣您就可以將其用於JS(if ajax)或php進行更靈活的處理
$csk = array();
while ($row = mysql_fetch_assoc($query)) {
$csk[] = array($row['NAME']);
}
echo $csk; //for ajax use echo json_encode($csk);
只是測試
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 $csk;
您將在每次循環迭代時將變量重置為$row['NAME']
的值。
您需要做的是將變量附加到$csk
:
$csk .= "'".$row['NAME']."',";
^---- notice the extra . here
額外的.
表示您要將值附加到$csk
。
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