我从我的PHP应用程序中得到以下警告,我想知道如何修复它:
警告:mysqli_fetch_array()希望参数1是mysqli_结果,布尔值在第59行的D:\WebShare\Students\Term\3Winter\PHP\5335-40\RosenthalD\enterproferences.PHP中给出 代码(第59行为粗体,以WHILE开头:我从我的PHP应用程序中得到以下警告,我想知道如何修复它:,php,mysql,error-handling,mysqli,Php,Mysql,Error Handling,Mysqli,警告:mysqli_fetch_array()希望参数1是mysqli_结果,布尔值在第59行的D:\WebShare\Students\Term\3Winter\PHP\5335-40\RosenthalD\enterproferences.PHP中给出 代码(第59行为粗体,以WHILE开头: $query="SELECT NewUsedID FROM zlu_newused WHERE CustomerID=" . $_SESSION['CustomerID2Update']
$query="SELECT NewUsedID FROM zlu_newused
WHERE CustomerID=" . $_SESSION['CustomerID2Update'] ; // Build the query
$rs = @mysqli_query ($dbc, $query); // Return the Result Set
WHILE ($row = mysqli_fetch_array($rs, MYSQLI_ASSOC)) { // Fetch the data
$NewUsedID = $row['NewUsedID'];
switch ($NewUsedID) {
case 1:
$rdoneworusedChecked1 = "CHECKED" ;
break;
case 2:
$rdoneworusedChecked2 = "CHECKED" ;
break;
case 3:
$rdoneworusedChecked3 = "CHECKED" ;
break;
}
}
如我所见,您的查询没有成功,请删除该@和。首先,不要使用
@
运算符,这是个坏主意
您的查询可能有错误。例如:
$query="SELECT NewUsedID FROM zlu_newused
WHERE CustomerID=" . $_SESSION['CustomerID2Update'] ; // Build the query
$rs = mysqli_query ($dbc, $query);
if(!$rs){
print mysqli_error($dbc);
}
将帮助您调试。这可能意味着查询因某种原因失败。在这一行中:
$rs = @mysqli_query ($dbc, $query);
删除用于抑制错误消息的@:
$rs = mysqli_query ($dbc, $query);
然后重试,看看是否收到错误消息并从那里开始工作