mysql沒有顯示錯誤訊息但是回傳的頁面是空白

edited 九月 2014 in 資料庫
<?php
header('content-type:text/html;charset=utf-8');
$dbServer = "localhost";
$dbName="shoping";
$dbPassworld="123";
$db="shop";

if(!mysqli_connect("$dbServer","$dbName","$dbPassworld","$db"))
{
 die("Fail");
}
else
{
 echo"connected";
}
$con = mysqli_connect("$dbServer","$dbName","$dbPassworld","$db");
mysqli_query($con,"SET NAMES utf8"); 
mysqli_query($con,"SET CHARACTER_SET utf8");
?>
<body>
<form name="form1" action="" method="post" enctype="multipart/form-date">
店家名稱:<input type="text"  name="store_name" size="5" >
店家電話:<input type="text" name="store_phone" size="10">
店家住址:<input type="text" name="store_live" size="30">
<input type="submit" value="新增">
<input type="reset" value="重設">
</form>
<?php
$name = $_POST['store_name'];
$live = $_POST['store_live'];
$phone = $_POST['store_phone'];

if(!isset($name)&&empty($name))
  if(!isset($live)&&empty($live))
   if(!isset($phone)&&empty($phone))
   {
     $sql = "INSERT shoping(live,name,phone)
	        value('$live,$name,$phone')
			";
			mysqli_query($con,$sql);
   }
   else
   {
    echo('<script language="JavaScript"
	      alert("資料表不可以有空白");
		>');
   }
   $sql = "SELECT * FROM shoping";
   
?>


<?php
   $result = mysqli_query($con,$sql);
  
   if(mysqli_num_rows($result)>0){
   echo"<table border='1' width='100%' >
   <tr>
   <td><center>店家名稱</td>
   <td><center>店家電話</td>
   <td><center>店家住址</td>
   </tr>";

while($row=mysqli_fetch_array($result))
{
echo"<tr>
          <td>{$row['name']}</td>
          <td>{$row['phone']}</td>
		  <td>{$row['live']}</td>
		  </tr>";
}
echo"</table>";
}
?>


</body>
我資料表都建立啊!,為什麼回傳都空白
https://scontent-b-hkg.xx.fbcdn.net/hphotos-xpf1/v/t1.0-9/10710666_1458047284463997_5091653026688734876_n.jpg?oh=1abe793855d148c067c9206f4e36cad0&oe=548680CB <-著是我建立的資料庫Myadmin

評論

  • 我後面追加
    $sql = "INSERT shoping(live,name,phone)
    value('$live,$name,$phone')
    ";
    $result = mysqli_query($con,$sql);
    if(!$result)
    {
    die('Invalid query: ' . mysql_error());
    }
    else
    {
    echo"ok";
    }
    他最後顯示Invalid query:失敗訊息.
    (努力找錯誤中)
  • 我也確認資料庫是否有連線
    又追加指令
    if(mysqli_select_db($con,"$db"))
    {
    echo"ok";
    }
    else
    {
    echo"faile";
    }
    他回傳ok
    資料庫已經連線了
  • <?php
    header('content-type:text/html;charset=utf-8');
    $dbServer = "localhost";
    $dbName="shoping";
    $dbPassworld="123";
    $db="shop";
    
    if(!mysqli_connect("$dbServer","$dbName","$dbPassworld","$db"))
    {
     die("Fail");
    }
    else
    {
     echo"connected";
    }
    echo"<br>";
    
    $con = mysqli_connect("$dbServer","$dbName","$dbPassworld","$db");
    mysqli_query($con,"SET NAMES utf8"); 
    mysqli_query($con,"SET CHARACTER_SET utf8");
    if(mysqli_select_db($con,$db))
    {
    echo"ok";
    }
    else
    {
    echo"fail";
    }
    error_reporting(E_ALL ^ E_DEPRECATED);
    ?>
    <body>
    <form name="form1" action="" method="post" enctype="multipart/form-date">
    店家名稱:<input type="text"  name="store_name" size="5" >
    店家電話:<input type="text" name="store_phone" size="10">
    店家住址:<input type="text" name="store_live" size="30">
    <input type="submit" value="新增">
    <input type="reset" value="重設">
    </form>
    <?php
    $name = $_POST['store_name'];
    $live = $_POST['store_live'];
    $phone = $_POST['store_phone'];
    
    if(!isset($name)&&empty($name))
      if(!isset($live)&&empty($live))
       if(!isset($phone)&&empty($phone))
       {
         $sql = "INSERT shoping(live,name,phone)
    	        value('$live,$name,$phone')
    			";
    		}	
       else
       {
        echo"資料表不可以有空白";
    		
       }
       $sql = "SELECT * FROM shoping";
       
    ?>
    
    
    <?php
       $result = mysqli_query($con,$sql);
    
      
       if(mysqli_num_rows($result)>0){
       echo"<table border='1' width='100%' >
       <tr>
       <td><center>店家名稱</td>
       <td><center>店家電話</td>
       <td><center>店家住址</td>
       </tr>";
    
    while($row=mysqli_fetch_array($result))
    {
    echo"<tr>
              <td>{$row['name']}</td>
              <td>{$row['phone']}</td>
    		  <td>{$row['live']}</td>
    		  </tr>";
    }
    echo"</table>";
    }
      $sql = "INSERT shoping(live,name,phone)
    	        value($live,$name,$phone)
    			";
    			$result = mysqli_query($con,$sql);
    			if(!$result)
    			{
    			die('Invalid query: ' . mysql_error());
    			}
    			else
    			{
    			echo"ok";
    			}
                
    ?>
    
    
    </body>
    
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