PHP/MYSQL使用datediff顯示mysql記錄的年齡

我是PHP/MYSQL的新手，我試圖從我的數據庫中使用DATEIFF顯示所有記錄的年齡，如我的代碼的第17行所示，但它只是將無法工作。我需要有人幫助我解決這個問題。PHP/MYSQL使用datediff顯示mysql記錄的年齡

$result = mysqli_query($con,"SELECT * FROM growers"); 

echo "<table class='table table-striped table-advance table-hover'> 
    <tbody> 
     <tr> 
      <th><i class='icon_profile'></i>&nbsp;Batch</th> 
      <th><i class='icon_ol'></i>&nbsp;Date Received</th> 
      <th><i class='icon_clock_alt'></i>&nbsp;Age when Received</th> 
      <th><i class='icon_clock_alt'></i>&nbsp;Current Age</th> 
      <th><i class='icon_star'></i>&nbsp;NO of Birds</th> 
      <th><i class='icon_info'></i>&nbsp;View More</th> 
     </tr>"; 

     while($row = mysqli_fetch_array($result)) 
     { 
      echo"<tr>"; 
      echo"<td>" . $row['BATCH'] . "</td>"; 
      echo"<td>" . $row['BIRTH DAY'] . "</td>"; 
      echo"<td>" . $row['AGE'] . "&nbsp;Week(s)" . "</td>"; 
      echo"<td>" . "SELECT DATEDIFF("NOW()", "$row['BIRTH DAY']") AS CURRENT AGE". "</td>"; 
      echo"<td>" . $row['NO OF BIRDS'] . "</td>"; 
      echo"<td>" . $row['AGE'] . "</td>"; 
      echo"</tr>"; 
     } 
echo "</table>"; 

mysqli_close($con); 
?>

來源

2016-11-13 Emmanuel Daniel

行'回聲「」。「SELECT DATEDIFF（」NOW（）「，」$ row ['BIRTH DAY']「）as CURRENT AGE」。「」'真的很混亂。您正試圖執行查詢並輸出結果，但實際上只顯示查詢。根據@ scaisEdge的回答建議，在第一行中計算查詢中的DATEDIFF（）。 – karliwson

你可以在主選擇選擇

$result = mysqli_query($con, 
     "SELECT 
     growers.BATCH, 
     growers.`BIRTH DAY`, 
     growers.AGE,    
     DATEDIFF(NOW(),growers.`BIRTH DAY`) AS CURRENT_AGE, 
     growers.`NO OF BIRDS`    
     FROM growers; " 
    ); 


         echo "<table class='table table-striped table-advance table-hover'> 
         <tbody> 
         <tr> 
         <th><i class='icon_profile'></i>&nbsp;Batch</th> 
         <th><i class='icon_ol'></i>&nbsp;Date Received</th> 
         <th><i class='icon_clock_alt'></i>&nbsp;Age when Received</th> 
         <th><i class='icon_clock_alt'></i>&nbsp;Current Age</th> 
         <th><i class='icon_star'></i>&nbsp;NO of Birds</th> 
         <th><i class='icon_info'></i>&nbsp;View More</th> 
         </tr>"; 

         while($row = mysqli_fetch_array($result)) 
         { 
         echo"<tr>"; 
         echo"<td>" . $row['BATCH'] . "</td>"; 
         echo"<td>" . $row['BIRTH DAY'] . "</td>"; 
         echo"<td>" . $row['AGE'] . "&nbsp;Week(s)" . "</td>"; 
         echo"<td>" . $row['CURRENT_AGE'] . "</td>"; 
         echo"<td>" . $row['NO OF BIRDS'] . "</td>"; 
         echo"<td>" . $row['AGE'] . "</td>"; 
         echo"</tr>"; 
         } 
         echo "</table>"; 

         mysqli_close($con); 
         ?>

來源

2016-11-13 21:43:26 scaisEdge

謝謝scaisEdge。但是我申請了這個建議後，我的桌子什麼都沒顯示。 –

沒有爲CURRENT_AGE或沒有爲所有的領域？ – scaisEdge

整個表消失了，除了標題 –

回答

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