Question

我制作了一个语音词典。用户输入一个单词并从我的sql数据库中获取相应的含义，它显示在屏幕上然后转换为语音。它工作正常。

我只想添加一个条件：如果用户输入了我的数据库中不存在的单词，那么它会输出消息＆＃34;抱歉的单词找不到＆＃34;，并且音频中不应该播放任何内容。目前，如果我输入一个不存在的单词，我将获得未定义索引的音频...等等...＆＃39;

请告诉我在哪里添加if条件和添加内容这是我的代码

<html>
<head>
<title>Word meanings</title> 
<?php

mysql_connect("localhost", "root", "abcd");
mysql_select_db("dictionary");
if(isset($_POST['Submit1']))

{

 $req=$_REQUEST['word'];
 $strSQL = "SELECT * FROM dict WHERE word='$req'";
 $rs = mysql_query($strSQL);

 while($row = mysql_fetch_array($rs))
 {
  $x=$row["word"];
  $y=$row["meaning"];
  $z=$row["synonym"];
  echo "<b>Word</b>: " . $x ."<br/>" ;
  echo "<b>Meaning</b>: " . $y ."<br/>" ;
  echo "<b>Synonym</b>: " . $z ."<br/>" ;
 }
}


mysql_close();
?>

</head>
<body>
<form name="form1" action="lastry.php" method="POST">
<input type="text" name="word" value="<?php echo isset($_POST['word'])?$_POST['word']:''?>"x-webkit-speech/>
<Input Type ="Submit" Name ="Submit1" Value ="submit"> 
</form>
<?php  
if($_POST)
{
?>

<p>Listen word 
<audio controls="controls">
<source src="http://speechutil.com/convert/ogg?text='<?php echo urlencode($x);?>'"    &type="audio/mp3" />

</audio>
</p>
<br><br>

<p>Listen meaning 
<audio controls="controls">


<source src="http://speechutil.com/convert/ogg?text='<?php echo urlencode($y);?>" &type="audio/mp3" />
</audio>
</p>
<br><br>
<p>Listen synonym
<audio controls="controls">

<source src="http://speechutil.com/convert/ogg?text='<?php echo urlencode($z);?>'" &type="audio/mp3" />
</audio>
</p>
<?php 
}

?>
</body>
</html>

Answer 1

也许检查它是否有价值

 if( isset($_POST) && !empty($x) )

Answer 2

在此处添加计数器：

 $foundwords = 0;    
 while($row = mysql_fetch_array($rs))
 {
  $x=$row["word"];
  $y=$row["meaning"];
  $z=$row["synonym"];
  echo "<b>Word</b>: " . $x ."<br/>" ;
  echo "<b>Meaning</b>: " . $y ."<br/>" ;
  echo "<b>Synonym</b>: " . $z ."<br/>" ;
  $foundwords++;  // increase counter after each found word
 }

然后在mysql_close()之后检查该变量（身体内某处）

if($foundwords==0) { 
  echo 'sorry word not found';
  ... finish html etc ...
  die();
}

Answer 3

在$rs = mysql_query($strSQL);之后，立即添加第$num_rows = mysql_num_rows($rs);行，然后您可以：

if($num_rows > 0) {
    /*Run while loop which you already have*/
}
else {
    echo 'Sorry word not found';
}

Answer 4

你应该这样做：

<?php
   if (isset($_POST['Submit1']) {
      mysql_connect("localhost", "root", "abcd");
      mysql_select_db("dictionary");

      $req=$_REQUEST['word'];
      $rs = mysql_query("SELECT * FROM dict WHERE word='$req'");

     while($row = mysql_fetch_array($rs)){
        $x=$row["word"];
        $y=$row["meaning"];
        $z=$row["synonym"];
        echo "<b>Word</b>: " . $x ."<br/>" ;
        echo "<b>Meaning</b>: " . $y ."<br/>" ;
        echo "<b>Synonym</b>: " . $z ."<br/>" ;
     }
     if (!isset($x)) { $All_Find = $req; } else { $All_Find = true; }

   }
?>

<?php if ($All_Find == true) { ?>
<audio controls="controls">
<source src="http://speechutil.com/convert/ogg?text='<?php echo urlencode($y);?>" &type="audio/mp3" />
</audio>
<?php } else { echo "Sorry, the world $All_Find cannot be found :(";}?>

Answer 5

未定义的索引消息主要是php.ini设置的结果。如果您关闭通知是警告错误级别。 php中不存在的关联数组成员的值始终为NULL（= false =＆＃39;＆＃39;）。

但是如果你想要确切，你可以像这样检查

// the condition evaluates to false if $row has no 'meaning' member
if($row['meaning'])
   $z = $row['meaning'];
else
   $z = "sorry word not found";

如果条件放入PHP代码怎么办？

5 个答案: