Question

我不知道PHP编码我设法将一些代码放在一起，从MySQL中提取一些数据我可以将结果看作HTML表格。

我无法弄清楚：
我想添加一个if语句，如if $x1 = ""然后不回显任何内容，否则返回html表中的结果。

这就是我所拥有的：

<?php 
$x1 = get_field(test);

$username = "xxxxx";
$password = "xxxxxx";
$hostname = "localhost"; 

//connection to the database
$dbhandle = mysql_connect($hostname, $username, $password) 
 or die("Unable to connect to MySQL");

// select a database to work with
$selected = mysql_select_db("xxxx",$dbhandle) 
  or die("Could not select examples");

// execute the SQL query and return records
$result = mysql_query("SELECT * FROM xxx WHERE xxxx = '$x1'");

// fetch tha data from the database
echo "<table><tr><th>Name</th><th>Nick Name</th><th>Email</th></tr>";

while ($row = mysql_fetch_array($result)) {
   echo 
"<tr><td>".$row["name_l"]."</td><td>".$row["nick_name"]."</td><td>".$row["email_s"]."</td></tr>";
}
echo "</table>";

//close the connection
mysql_close($dbhandle);

上面代码的问题是当$x1 = ""时，会回显标题：/

Answer 1

$x1 = get_field(test);

if($x1!=""){
$username = "xxxxx";
$password = "xxxxxx";
$hostname = "localhost"; 

//connection to the database
$dbhandle = mysql_connect($hostname, $username, $password) 
 or die("Unable to connect to MySQL");

//select a database to work with
$selected = mysql_select_db("xxxx",$dbhandle) 
  or die("Could not select examples");

//execute the SQL query and return records
$result = mysql_query("SELECT * FROM xxx WHERE xxxx = '$x1'");

//fetch tha data from the database

echo "<table>
<tr><th>Name</th><th>Nick Name</th><th>Email</th></tr>";
while ($row = mysql_fetch_array($result)) {
   echo 
"<tr><td>".$row["name_l"]."</td><td>".$row["nick_name"]."</td><td>".$row["email_s"]."</td></tr>";
}
echo "</table>";

//close the connection
mysql_close($dbhandle);}
?>

Answer 2

用以下代码替换您的代码：

<?php 
$x1 = get_field("test");
if($x1){

$username = "xxxxx";
$password = "xxxxxx";
$hostname = "localhost"; 

//connection to the database
$dbhandle = mysql_connect($hostname, $username, $password) 
or die("Unable to connect to MySQL");

//select a database to work with
$selected = mysql_select_db("xxxx",$dbhandle) 
or die("Could not select examples");

//execute the SQL query and return records
$result = mysql_query("SELECT * FROM xxx WHERE xxxx = '$x1'");

//fetch tha data from the database

echo "<table>
<tr><th>Name</th><th>Nick Name</th><th>Email</th></tr>";
while ($row = mysql_fetch_array($result)) {
echo 
"<tr><td>".$row["name_l"]."</td><td>".$row["nick_name"]."   </td><td>".$row["email_s"]."</td></tr>";
}
echo "</table>";

//close the connection
mysql_close($dbhandle);
} 
else {echo "Sorry, Nothing to display"; }

?>

Answer 3

好的，我是如何解决的。

//fetch tha data from the database
if ($x1 == '') {
echo "<table>
<tr><th>Name</th><th>Nick Name</th><th>Email</th></tr>";
while ($row = mysql_fetch_array($result)) {
   echo 
"<tr><td>".$row["name_l"]."</td><td>".$row["nick_name"]."</td><td>".$row["email_s"]."</td></tr>";
}
echo "</table>";} else {}

//close the connection
mysql_close($dbhandle);

?>

PHP - 在这种情况下如何实现IF？

3 个答案: