當沒有找到記錄時不顯示消息

Question

由於某種原因，當查詢數據庫的選定範圍時，下面沒有顯示「找不到結果」錯誤消息，但正在顯示錶頭和頁腳。

$result = mysqli_query($con,"SELECT * FROM tblRecords WHERE DATE(RecDate) = CURDATE() - INTERVAL 1 DAY ORDER BY RecDate DESC, RecTime DESC"); 

<?php 

if (!$result) { echo("No results found for the selected view"); 
      } else ?> 
      <table id="results"> 
<tr> 
<th>Rec#</th> 
<th>Date</th> 
<th>Time</th> 
<th>Reading</th> 
</tr> 
<?php ; 
while($row = mysqli_fetch_array($result)) 
?> 
<tr> 

<td><?php echo($row['RecID']);?></td> 
<td><?php echo(date("d/m/Y", strtotime($row['RecDate'])));?></td> 
<td><?php echo(date("g:i A", strtotime($row['RecTime'])));?></td> 
<td><?php echo($row['RecReading'] . $row['RecMeasure']);?></td> 
</tr> 
<?php 
} 
?> 
<tr> 
<td class="footer" colspan="4">- end of report -</td></tr> 
</table> 
<?php 
mysqli_close($con); 
?> 

任何援助將不勝感激理論上，這應該工作......不應該嗎？ :-)

Answer 1

你可以用這個條件來檢查查詢行號檢查

$result = mysqli_query($con,"SELECT * FROM tblRecords WHERE DATE(RecDate) = CURDATE() - INTERVAL 1 DAY ORDER BY RecDate DESC, RecTime DESC"); 

<?php 

if (mysqli_num_rows($result) === 0) { 
    echo("No results found for the selected view"); 
} else {?> 
<table id="results"> 
<tr> 
<th>Rec#</th> 
<th>Date</th> 
<th>Time</th> 
<th>Reading</th> 
</tr> 
<?php ; 
while($row = mysqli_fetch_array($result)) 
?> 
<tr> 

<td><?php echo($row['RecID']);?></td> 
<td><?php echo(date("d/m/Y", strtotime($row['RecDate'])));?></td> 
<td><?php echo(date("g:i A", strtotime($row['RecTime'])));?></td> 
<td><?php echo($row['RecReading'] . $row['RecMeasure']);?></td> 
</tr> 
<?php } ?> 
<tr> 
<td class="footer" colspan="4">- end of report -</td></tr> 
</table> 
<?php 
mysqli_close($con); 
?> 

Answer 2

$result可能是一個結果集，但它可能是空的。但!$result將不會成立。 mysql_query的文檔：

返回FALSE失敗。爲成功SELECT,SHOW,DESCRIBE或EXPLAIN查詢mysqli_query()將返回一個mysqli_result對象。對於其他成功的查詢mysqli_query()將返回TRUE。 （來源：http://php.net/mysqli_query）

你應該mysqli_num_rows（或類似的東西）