[英]echo something when no results where found using sql LIKE
我在我的網站上做了一個搜索框,搜索效果很好。 但是,當我輸入數據庫(在我要搜索的數據庫)中找不到的隨機字符串時,將顯示空白頁面。 如何顯示類似“找不到匹配項”的內容? 這是我正在使用的代碼:
$search = $_GET["zoek"];
$result = mysqli_query($con,"SELECT Afbeelding,Product,Prijs,Beschrijving FROM Producten WHERE Product LIKE '%$search%' order by Product ASC LIMIT 0, 5");
echo '<table border="1px solid black" cellspacing="0" style="margin-top:47px"><tbody>';
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td rowspan='2' width= '200'><img src='" . $row['Afbeelding'] . "' width= '200' height='250'></td>";
echo "<td><b>" . $row['Product'] . "</b></td>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row['Beschrijving'] . " <i>Prijs: € " . $row['Prijs'] . "</i><br/><br/></td>";
echo "</tr>";
}
echo "</tbody></table>";
只是問一下,該查詢通過mysqli_num_rows()返回多少結果:
$result = mysqli_query($con,"SELECT Afbeelding,Product,Prijs,Beschrijving FROM Producten WHERE Product LIKE '%$search%' order by Product ASC LIMIT 0, 5");
if(mysqli_num_rows($result) > 0) {
echo '<table border="1px solid black" cellspacing="0" style="margin-top:47px"><tbody>';
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td rowspan='2' width= '200'><img src='" . $row['Afbeelding'] . "' width= '200' height='250'></td>";
echo "<td><b>" . $row['Product'] . "</b></td>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row['Beschrijving'] . " <i>Prijs: € " . $row['Prijs'] . "</i><br/><br/></td>";
echo "</tr>";
}
echo "</tbody></table>";
} else {
echo "No matches found";
}
試試這個。
$search = $_GET["zoek"];
$result = mysqli_query($con,"SELECT Afbeelding,Product,Prijs,Beschrijving FROM Producten WHERE Product LIKE '%$search%' order by Product ASC LIMIT 0, 5");
$row_cnt = mysqli_num_rows($result);
if($row_cnt>0)
{
echo '<table border="1px solid black" cellspacing="0" style="margin-top:47px"><tbody>';
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td rowspan='2' width= '200'><img src='" . $row['Afbeelding'] . "' width= '200' height='250'></td>";
echo "<td><b>" . $row['Product'] . "</b></td>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row['Beschrijving'] . " <i>Prijs: € " . $row['Prijs'] . "</i><br/><br/></td>";
echo "</tr>";
}
echo "</tbody></table>";
}
else
{
echo "Sorry. No Results!";
}
您應該嘗試一下。
$search = $_GET["zoek"];
$result = mysqli_query($con,"SELECT Afbeelding,Product,Prijs,Beschrijving FROM Producten
WHERE Product LIKE '%$search%' order by Product ASC LIMIT 0, 5");
$total_rows=mysqli_fetch_row($result);
if($total_rows>0){
echo '<table border="1px solid black" cellspacing="0" style="margin-top:47px"><tbody>';
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td rowspan='2' width= '200'><img src='" . $row['Afbeelding'] . "' width= '200' height='250'></td>";
echo "<td><b>" . $row['Product'] . "</b></td>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row['Beschrijving'] . " <i>Prijs: € " . $row['Prijs'] . "</i><br/><br/></td>";
echo "</tr>";
}
echo "</tbody></table>";
}
else{
echo "NO match found";
}
Thanks
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