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好的，我几乎从我的数据库中获得了一个下拉菜单并发布到它以工作。我让它拉下数据并将其提交给数据库。还是一个树桩。如果我有示例“ABC Trucking”作为选项。它只将“ABC”发布到 table1。无论出于何种原因，它都没有发布两个词？有任何想法吗？查看运营商名称下拉菜单在 div 中的位置。

我的代码：

<?php

if (isset($_POST["submit"]) && $_POST["submit"] == "Submit")
{
    for ($count = 1; $count <= 9; $count++)
    {
        $fields[$count] = "";
        if (isset($_POST["field" . $count . ""]))
        {
            $fields[$count] = trim($_POST["field" . $count . ""]);
            //echo $fields[$count] . "<br />";
        }
    }

    $con = mysql_connect("local", "user", "pass");
    mysql_select_db("DB", $con);

    $carriername = mysql_real_escape_string($_POST['carriername']);
    $fromzip = mysql_real_escape_string($_POST['fromzip']);
    $tozip = mysql_real_escape_string($_POST['tozip']);
    $typeofequipment = mysql_real_escape_string($_POST['typeofequipment']);
    $weight = mysql_real_escape_string($_POST['weight']);
    $length = mysql_real_escape_string($_POST['length']);
    $paymentamount = mysql_real_escape_string($_POST['paymentamount']);
    $contactperson = mysql_real_escape_string($_POST['contactperson']);
    $loadtype = mysql_real_escape_string($_POST['loadtype']);

    $insert = "INSERT INTO table1 (`carriername` ,`fromzip` ,`tozip` ,`typeofequipment` ,`weight` ,`length` ,`paymentamount` ,`contactperson` ,`loadtype`) VALUES('$carriername' ,'$fromzip' ,'$tozip' ,'$typeofequipment' ,'$weight' ,'$length' ,'$paymentamount' ,'$contactperson' ,'$loadtype');";
    mysql_query($insert) or die(mysql_error());

    $select = "SELECT `carriername` ,`fromzip` ,`tozip` ,`typeofequipment` ,`weight` ,`length` ,`paymentamount` ,`contactperson` ,`loadtype` FROM `table1` ORDER BY `paymentamount` DESC;";
    $result = mysql_query($select) or die(mysql_error());
}
?>
</script>
<style ="text-align: center; margin-left: auto; margin-right: auto;"></style>
</head>
<body>
<div
 style="border: 2px solid rgb(0, 0, 0); margin: 16px 20px 20px; width: 400px; background-color: rgb(236, 233, 216); text-align: center; float: left;">
<form action="" method="post";">
  <div
  style="margin: 8px auto auto; width: 300px; font-family: arial; text-align: left;"><br>
  <table style="font-weight: normal; width: 100%; font-size: 12px;"
 border="1" bordercolor="#929087" cellpadding="6" cellspacing="0">
   <table
 style="font-weight: normal; width: 100%; text-align: right; font-size: 12px;"
 border="1" bordercolor="#929087" cellpadding="6" cellspacing="0">
    <tbody>
    <tr>
    <td style="width: 10%;">Carrier:</td><td>
 <?php  

$con = mysql_connect("local", "user", "pass");
    mysql_select_db("DB", $con);

$query=("SELECT * FROM table2"); 

$result=mysql_query($query) or die ("Unable to Make the Query:" . mysql_error() ); 
echo "<select name=carriername>";   

while($row=mysql_fetch_array($result)){  

echo "<OPTION VALUE=".$row['carriername'].">".$row['carriername']."</OPTION>"; 
} 

echo "</select>";  


?>
              </td>
        </tr>
        <tr>
              <td style="width: 35%;">Pick Zip:</td><td> <input id="fromzip" name="fromzip" maxlength="50"
 style="width: 100%;" type="text">
        </tr>
        <tr>
              <td style="width: 35%;">Drop Zip:</td><td> <input id="tozip" name="tozip" maxlength="50"
 style="width: 100%;" type="text">
        </tr>
        <tr>
              <td style="width: 35%;">Load Type:</td><td> <input id="loadtype" name="loadtype" maxlength="50"
 style="width: 100%;" type="text">
        </tr>
        <tr>
              <td style="width: 35%;">Rate:</td><td> <input id="paymentamount" name="paymentamount" maxlength="50"
 style="width: 100%;" type="text">
        </tr>
          </tbody>
  </table>
  <p style="text-align: center;"><input name="submit" value="Submit"
 class="submit" type="submit"></p>
  </div>
</form>
</div>
<p style="margin-bottom: -20px;">&nbsp;</p>
</body>
4

1 回答 1

2

代替 ：

echo "<OPTION VALUE=".$row['carriername'].">".$row['carriername']."</OPTION>";

用这个

echo "<OPTION VALUE='".$row['carriername']."'>".$row['carriername']."</OPTION>";

注意' in ur value...concate ''到你的 value attr .. 使它成为一个字符串....

已编辑

echo "<select name='carriername'>";   
while($row=mysql_fetch_array($result)){  
  echo "<OPTION VALUE=".$row['carriername'].">".$row['carriername']."</OPTION>"; 
} 
echo "</select>";
于 2012-11-06T07:57:06.170 回答