douzhuanqian8244 2013-04-16 14:47 采纳率: 100%
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使用数组填充选择输入

I would like to start out saying I don't want to use Javascript....

I have an array which is returning correctly I believe, however I am only having the first entry in the array display in the drop down list.

Is this a problem with the array? or with this function?

foreach($array as $key=>$value){
    $html = "<option value='$key'>$value</key>";
}

echo "<select name="process">$html</select>";
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  • dongyu3712 2013-04-16 14:49
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    You have to use the concatenation operator (.):

    $html = '';
    foreach($array as $key => $value)
    {
        $html.= "<option value='$key'>$value</option>";
    }
    
    echo "<select name=\"process\">$html</select>";
    

    However, looking at the function you posted earlier, mysql_fetch_assoc only returns a single row at a time. You need to loop over that, instead. The following should suffice:

    function tasks_list($p_id) { 
        $project_tasks = array(); 
        $p_id = (int)
        $p_id; 
        $func_num_args = func_num_args(); 
        $func_get_args = func_get_args(); 
    
        $result = mysql_query("SELECT task_name FROM tasks WHERE project_id = $p_id");
        while($project_tasks = mysql_fetch_assoc($result))
        {
            $html.= "<option value='".$project_tasks['task_name']."'>".$project_tasks['task_name']."</option>";
        }
    
        echo "<select name=\"process\">$html</select>";
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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