PHP printf函数“提供的参数不是有效的MySQL结果资源” [重复]

Question

您好，我的以下代码在一台服务器上运行良好，但在我使用的免费服务器上出现以下错误

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/a9645958/public_html/tteeth/staffslot.php on line 75

我的代码如下：

// Starting of top line showing name of the day's slots
echo "<form method='post' name='time' action='timeinsert.php'>
<table align='center' width='380' border='1'>
<tr>
<th>Select</th>
<th>Appointment Times</th>
</tr>";

// Starting of the slots
for($i=0;$i<=31;$i++){ // this shows 32 x 15 minute slots from 9:00am - 17:00pm
$shr=9+floor($i/4); //calculates the start hour of a slot
$smin=($i%4)*15; //calculates the start minutes of a slot
$time=mktime($shr,$smin,00,0,0,0); //creates full php time from hours and minutes above
$displayTime2=date("H:i:s",$time); // creates correct time format for MySQL database hrs, mins, seconds

$pagetime = date("H:i a", strtotime($displayTime2)); //converts the time to just hours and minutes
$timenow = date('H:i'); // get's the current time

$chosendate = date(mktime(0,0,0,$month,$day,$year)); // converts the date picked to php time (seconds since 01 01 1971)
$todayDate = date(mktime()); // converts today's date to php time so it can be compared with the chosen date

while ($myrow = mysql_fetch_row($result)) {
        printf("<tr><td>%s</td><td>%s</td></tr>\n",
        $myrow[0]);
}
$result=mysql_query("SELECT * FROM appointment where Appointment_Time='$displayTime2' AND Appointment_Date='$insertdate' AND Practice_ID='$practice'");
$myrow = mysql_fetch_row($result);
    if($pagetime<$timenow AND $chosendate<$todayDate) //if the display time is less than the time now, and the date is less than todays date
        { //do not display a row 
        }
    elseif($myrow)// if the values in the database match the row slot time
        {
        echo "<td></td><td align='center'>$pagetime</td>";//just display the slot as a time
        }
    else
        {
        echo    "<td align='center'><input type='checkbox' name='$displayTime2' onclick='KeepCount()'></td>
                <td align='center'>$pagetime</td>";
        }
    echo "</td></tr>";          
        }
    echo "</table><br>
        <input type='submit' value='Submit'>
</form>";

我很确定问题出在 print f 函数上，因为它要求为 $myrow 中的值打印表头，但它还没有得到它的值...但是如果我将它移到 select 语句之后，它会弄乱我的其余代码（它不会将时间与数据库中的内容进行比较，并且在下一页上插入数据库时​​存在问题）

Answer 1

在这里，您正在尝试获取当时不存在的结果。

while ($myrow = mysql_fetch_row($result)) {
        printf("<tr><td>%s</td><td>%s</td></tr>\n",
        $myrow[0]);
}

你应该把那个while循环放在这个之后

$result=mysql_query("SELECT * FROM appointment where Appointment_Time='$displayTime2' AND Appointment_Date='$insertdate' AND Practice_ID='$practice'");