使用MYSQL和下拉列表

好吧，我幾乎從我的數據庫中拉出了一個下拉列表，併發布到它上面以及工作。我得到它來提取數據並將其提交給數據庫。還是一個殘端。如果我有一個示例「ABC卡車」作爲選項。它只將「ABC」發佈到table1。無論出於何種原因，它可以發佈兩個詞？有任何想法嗎？查看carriername下拉列表在div中的位置。使用MYSQL和下拉列表

我的代碼：

<?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>

來源

2012-11-06 user1801996

代替：

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

使用本

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

通知「在烏爾值 ... concate '您值attr ..所以它使它成爲一個字符串....

EDITED

echo "<select name='carriername'>"; 
while($row=mysql_fetch_array($result)){ 
    echo "<OPTION VALUE=".$row['carriername'].">".$row['carriername']."</OPTION>"; 
} 
echo "</select>";

來源

2012-11-06 07:57:06 bipen

有了它現在犯規讀取所有的運營商。它回到了唯一的選擇顯示「布什貨運」像這樣「Abc卡車運輸> Abc卡車」 – user1801996

不讓你...更新的答案，請檢查.. – bipen

Aslo我複製你的方式，你期待一個;在線上。 – user1801996

使用MYSQL和下拉列表

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