與mySQL查詢有關的麻煩。需要*而不是'*'

Question

我有以下代碼不起作用。這是行不通的，因爲它會做一個查詢

WHERE列=「*」，而不是WHERE列= *

我試圖想辦法來獲得它，所以它會做WHERE變量='var'如果一個變量在表單中發佈，而WHERE column = *如果沒有發佈，但我想不出一種方式，而我嘗試的所有內容都是hacky或不工作。

if(isset($_POST['variable'])){ 
    $variable=$_POST['variable']; 
    } 
    else{$variable='*';} 
    $sql="SELECT * FROM table WHERE column = '$variable'"; 

編輯，下面是實際的代碼：

<form method='post' action='policy.php?go'> 
<input type='radio' name='gen' value='M'>Male 
<input type='radio' name='gen' value='F'>Female 
<select name='state'> 
<option value='AK'>AK</option> 
<option value='WY'>WY</option> 
</select> 
<input type='radio' name='logic' value='>'>Older Than 
<input type='radio' name='logic' value='<'>Younger Than 
<select name='age'> 
<option value='5'>5</option> 
<option value='11'>11</option> 
<option value='17'>17</option> 
<option value='65'>65</option> 
</select> 
<input type='submit' name='submit' value='Search'> 
</form> 
<?php 
if(isset($_GET['go']) && isset($_POST['submit'])){ 
if(isset($_POST['state'])){ 
$state="'".mysql_real_escape_string($_POST['state'])."'"; 
} 
else{ $state='*';} 
if(isset($_POST['age'])){ 
$age=$_POST['age']; 
//append to query string 
} 
if(isset($_POST['logic'])){ 
$log=$_POST['logic']; 
//append to query string 
} 
else{$log='';} 
if(isset($_POST['gen'])){ 
$gen="'".mysql_real_escape_string($_POST['gen'])."'"; 
} 
else {$gen='*';} 
echo "<table id='hor-minimalist-b' summary='Employee Pay Sheet' class='tablesorter'> 
<caption>Age: ".$log." ".$age." Gender: ".$gen." </caption> 
<thead> 
<tr> 
<th scope='col'>State</th> 
<th scope='col'>Number</th> 
</tr> 
</thead> 
<tbody>"; 
// (WHY IS THIS NOT WORKING?) 
$sql = "SELECT SUM(num) AS sum, state,gen,age FROM `policy-ssi` 
WHERE age $log $age AND gen = $gen 
GROUP BY state 
ORDER BY sum DESC"; 
$result = mysql_query($sql); 
while ($row = mysql_fetch_assoc($result)) { 

Answer 1

在MySQL中，字符串的通配符是%而非*。爲了不管什麼在它的領域相匹配，改變else {$gen='*';}到else {$gen='\"%\"';}

此外，爲了使用通配符，您的查詢應該成爲

$sql = "SELECT SUM(num) AS sum, state,gen,age FROM `policy-ssi` 
WHERE age $log $age AND gen LIKE $gen 
GROUP BY state 
ORDER BY sum DESC";