如何从数据库中获取数据并将其显示在PHP中？

Question

如何使用php从数据库中获取数据并显示它？

数据库表具有列，标记为ID & Number。 ID是唯一的&固定，而数字只是一个非唯一的数字。如果有人访问 http://example.com/show.php?ID=32，并show.php应取Number &显示"Your number is XXX”

请提供代码样本。

Answer 1

<?php 
//Connect to DB 
$db = mysql_connect("localhst","user","pass") or die("Database Error"); 
mysql_select_db("db_name",$db); 

//Get ID from request 
$id = isset($_GET['id']) ? (int)$_GET['id'] : 0; 

//Check id is valid 
if($id > 0) 
{ 
    //Query the DB 
    $resource = mysql_query("SELECT * FROM table WHERE id = " . $id); 
    if($resource === false) 
    { 
     die("Database Error"); 
    } 

    if(mysql_num_rows($resource) == 0) 
    { 
     die("No User Exists"); 
    } 

    $user = mysql_fetch_assoc($resource); 

    echo "Hello User, your number is" . $user['number']; 
} 

这是非常基本的，但应该会看到你。

Answer 2

尝试

SELECT number from numberTable nt 
JOIN idTable it ON it.ID = nt.ID 
WHERE it.ID = `your given id` 

，因为我认为你的两个表与id

Answer 3

首先引用获取用户的ID（它可以在访问给定的或者基于在访问给定的细节）

然后将选择查询写入包含'number'字段的表格中。like

SELECT number FROM table1 WHERE table1.ID=IDFromtheuser; 

Answer 4

<?php 
$host="localhost"; 
$username=""; 
$password=""; 
$db_name="multiple_del"; 
$tbl_name="test_mysql"; 

// Connect to server and select databse. 
mysql_connect("$host", "root", "")or die("cannot connect"); 
mysql_select_db("multiple_del")or die("cannot select DB"); 

$sql="SELECT * FROM test_mysql"; 


$result=mysql_query($sql); 

$count=mysql_num_rows($result); 

?> 

<table width="400" border="0" cellspacing="1" cellpadding="0"> 
<tr> 
<td><form name="form1" method="post" action=""> 
<table width="400" border="0" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC"> 
<tr> 
<td bgcolor="#FFFFFF">&nbsp;</td> 
<td colspan="4" bgcolor="#FFFFFF"><strong>Delete multiple rows in mysql</strong> </td> 
</tr> 
<tr> 
<td align="center" bgcolor="#FFFFFF">#</td> 
<td align="center" bgcolor="#FFFFFF"><strong>Id</strong></td> 
<td align="center" bgcolor="#FFFFFF"><strong>Name</strong></td> 
<td align="center" bgcolor="#FFFFFF"><strong>Lastname</strong></td> 
<td align="center" bgcolor="#FFFFFF"><strong>Email</strong></td> 
</tr> 

<?php 
while($rows=mysql_fetch_array($result)){ 
?> 

<tr> 
<td align="center" bgcolor="#FFFFFF"><input name="checkbox[]" type="checkbox" id="checkbox[]" value="<? echo $rows['id']; ?>"></td> 
<td bgcolor="#FFFFFF"><? echo $rows['id']; ?></td> 
<td bgcolor="#FFFFFF"><? echo $rows['name']; ?></td> 
<td bgcolor="#FFFFFF"><? echo $rows['lastname']; ?></td> 
<td bgcolor="#FFFFFF"><? echo $rows['email']; ?></td> 
</tr> 

<?php 
} 
?> 

<tr> 
<td colspan="5" align="center" bgcolor="#FFFFFF"><input name="delete" type="submit" id="delete" value="Delete"></td> 
</tr> 

<?php 


if($delete){ 
for($i=0;$i<$count;$i++){ 
$del_id = $checkbox[$i]; 
$sql = "DELETE FROM test_mysql WHERE id='$del_id'"; 
$result = mysql_query($sql); 
} 


} 

?> 

</table> 
</form> 
</td> 
</tr> 
</table> 



its my code but its not working and error are show in this code:- 

<?php 


if($delete){ 
for($i=0;$i<$count;$i++){ 
$del_id = $checkbox[$i]; 
$sql = "DELETE FROM test_mysql WHERE id='$del_id'"; 
$result = mysql_query($sql); 
} 


} 

?> 

please help me.