回聲的東西，當沒有結果發現使用sql LIKE

我做了一個搜索框在我的網站和搜索工作正常。但是，當我輸入一個隨機字符串，在我的數據庫中找不到（我正在查找）一個空白頁面顯示。我如何顯示「找不到匹配」的內容？這是我使用的代碼：回聲的東西，當沒有結果發現使用sql LIKE

$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'] . "&nbsp;<i>Prijs: &euro;&nbsp;" . $row['Prijs'] . "</i><br/><br/></td>"; 
    echo "</tr>"; 

    } 

    echo "</tbody></table>";

來源

2014-01-28 EerlijkeDame

use if（）{} else {}。所以就像如果num_rows> 0顯示結果，否則回顯沒有結果 –

我試過如果其他人，只是不知道要使用什麼條件：P我會盡快嘗試這一點，當我回家！ – EerlijkeDame

查看另一個使用已發佈的答案，這就是我的原因:) –

就問，這與查詢mysqli_num_rows()多少結果返回：

$result = mysqli_query($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'] . "&nbsp;<i>Prijs: &euro;&nbsp;" . $row['Prijs'] . "</i><br/><br/></td>"; 
    echo "</tr>"; 

    } 

    echo "</tbody></table>"; 
} else { 
    echo "No matches found"; 
}

來源

2014-01-28 09:29:20

嘗試與此有關。

$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'] . "&nbsp;<i>Prijs: &euro;&nbsp;" . $row['Prijs'] . "</i><br/><br/></td>"; 
    echo "</tr>"; 

    } 

    echo "</tbody></table>"; 
    } 
    else 
    { 
    echo "Sorry. No Results!"; 
    }

來源

2014-01-28 09:30:28

你應該試試這個。

$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"); 
    $total_rows=mysqli_fetch_row($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'] . "&nbsp;<i>Prijs: &euro;&nbsp;" . $row['Prijs'] . "</i><br/><br/></td>"; 
     echo "</tr>"; 

     } 

     echo "</tbody></table>"; 
    } 
    else{ 
     echo "NO match found"; 
    } 
Thanks

來源

2014-01-28 09:41:19 user12

回聲的東西，當沒有結果發現使用sql LIKE

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