从列表中选择数据库，然后在PHP中查询它

Question

I would like to know what I'm missing in my code. 我想知道我的代码中缺少什么。

I have this HTML code: 我有这个HTML代码：

<html>
  <head>
    <title>Human gene catalog</title>
  </head>
<body>
  <h1>Reference sequence and Gene Ontology catalog</h1>
  <p>
    <b>1.Search for a gene:</b>
    <form action=prueba.php method=post>
      <select name=organism> 
        <option value=1>Human</option> 
        <option value=2>Mouse</option> 
        <option value=3>Zebrafish</option> 
        <option value=4>Fruit fly</option> 
      </select>
      <label>Please select a gene:</label>
      <br/>
      <input type=text name=gene>
      <br/><br/>
      <input type=submit name=submit value=Submit>
    </form>
  </p>
</html>

And this PHP code: 这个PHP代码：

 <?php
   $gene = $_POST["gene"];
   $specie = $_POST["organism"];

   $enlace = mysqli_connect("localhost","root","******","refGene");
   if (mysqli_connect_errno()) {
       echo "Failed to connect to MySQL: " . mysqli_connect_error();
   }

   if ($specie == 1) {
       mysqli_select_db($enlace,"refGene_human");
   } elseif ($specie == 2) {
       mysqli_select_db($enlace,"refGene_mouse");
   } elseif ($specie == 3) {
       mysqli_select_db($enlace,"refGene_zebrafish");
   } elseif ($specie == 4) {
       mysqly_select_db($enlace,"refGene_fruitfly");
   } else {
       echo "The gene is not in database";
   }

   $result = mysqli_query($enlace,"select * from (here I dont know what to put,if with one specie, here I put the name of the table) where name2 like '%$gene%'");

   echo "<h1>Gene Reference Results</h1>";

   echo "<table cellspacing=3 cellpadding=4 border=1 bgcolor=#dddddd>";
   echo "<tr align='center'><th>Transcript</th><th>Gene</th <th>Chromosome</th><th>Strand</th><th>Gene_Start</th><th>Gene_End</th><th>CDS_Start</th><th>CDS_End</th><th>ExonCount</th>";

   while ($extraido = mysqli_fetch_array($result)){
       echo "<tr>";
       echo "<td>".$extraido['name']."<br/>";
       echo "<td>".$extraido['name2']."<br/>";
       echo "<td align='center'>".$extraido['chrom']."<br/>";
       echo "<td align='center'>".$extraido['strand']."<br/>";
       echo "<td align='right'>".$extraido['txStart']."<br/>";
       echo "<td align='right'>".$extraido['txEnd']."<br/>";
       echo "<td align='right'>".$extraido['cdsStart']."<br/>";
       echo "<td align='right'>".$extraido['cdsEnd']."<br/>";
       echo "<td align='right'>".$extraido['exonCount']."<br/>";
   }
   echo "</table>";

   mysqli_free_result($result);
   mysqli_close($enlace);

Basically what I want to achieve is that, selecting a specie and a gene name in the HTML document, it gets you to the PHP, where, depending on the value of specie (1,2,3,4) it selects the corresponding database, and then querying it to find some information. 基本上，我想要实现的是，在HTML文档中选择一个物种和一个基因名称，它将带您进入PHP，在此，根据物种的值（1,2,3,4）选择相应的数据库，然后查询它以查找一些信息。

I have the right code to make it work but just with one specie, without the select list in the HTML and the if statements in the PHP, but I would like to make it work in more than one. 我有正确的代码可以使它工作，但是只需要一个种类，而HTML中没有选择列表，PHP中没有if语句，但是我想使它在多个中工作。

I think I should declare some variable before the if statement in the PHP and then querying over this variable which should tell the query function what table to choose. 我认为我应该在PHP的if语句之前声明一些变量，然后查询该变量，这应该告诉查询函数选择哪个表。

But I don't know what could be the right syntax. 但是我不知道什么是正确的语法。

Answer 1

if ($specie == 1) {
    mysqli_select_db($enlace,"refGene_human");
    $database = 'refGene_human';
} 
elseif ($specie == 2) {
    mysqli_select_db($enlace,"refGene_mouse");
    $database = 'refGene_mouse';
} 
elseif ($specie == 3) {
    mysqli_select_db($enlace,"refGene_zebrafish");
    $database = 'refGene_zebrafish';
} 
elseif ($specie == 4) {
    mysqly_select_db($enlace,"refGene_fruitfly");
    $database = 'refGene_fruitfly'; 
} 
else { 
    echo "The gene is not in database";
}

$result = mysqli_query($enlace,"select * from '$database' where name2 like '%$gene%'");

从列表中选择数据库，然后在PHP中查询它

问题描述

1 个解决方案

解决方案1
0 已采纳 2016-12-13 10:06:19

从列表中选择数据库，然后在PHP中查询它

问题描述

1 个解决方案

解决方案1 0 已采纳 2016-12-13 10:06:19

解决方案1
0 已采纳 2016-12-13 10:06:19