自動變化選項的值/ MYSQL

$MethodLocationFrom =mysql_query(" 
    SELECT DISTINCT Location_From AS 'locationFrom' 
    FROM trip 
"); 
    echo "<form action='confirmation.php' method='post'>"; 
    echo "<select class = 'LocationFrom' size='1' name='locationFrom' id='locationFrom'>"; 
    while($check = mysql_fetch_array($MethodLocationFrom)) 
{ 
echo "<option value='" . $check['locationFrom'] . "'>" . $check['locationFrom'] . "</option>"; 
} 
echo"</select>"; 

$MethodLocationTo =mysql_query(" 
    SELECT DISTINCT Location_To AS 'locationTo' 
    FROM trip 
"); 
    echo "<select class = 'LocationTo' size='1' size='1' name='locationTo' id='locationTo'>"; 
    while($check = mysql_fetch_array($MethodLocationTo)) 
{ 
    echo "<option value='" . $check['locationTo'] . "'>" . $check['locationTo'] . "</option>"; 
} 
echo"</select>";

output of the code 自動變化選項的值/ MYSQL

嗨，大家好！我正在制定航天飛機預訂系統，我遇到了選項框問題，我想要做的是當我從位置（從第一個選項框）中選擇一個位置時，該位置將自動在我想要的位置移除去（第二個箱子）我該怎麼做？被（減少簡潔）

來源

2013-11-24 Gaston Velarde

好吧，讓我一展身手，與個人的扭曲

假設迭代的結果

<select id="locationFrom"> 
    ... 
    <option value="$from">($from)</option> 
</select> 
<select id="locationTo"> 
    ... 
    <option value="$to">($to)</option> 
</select>

我認爲每個選項設定的值是相同。

讓我們用javascript來弄髒。請注意，這段時間很長，沒有任何評論，請查看它，測試它並在發現錯誤時給我反饋。

// script.js 
(function(){ 
    var selectFrom, selectTo, unsetChild, nextSibling, 
    onSelect, removeOption, recoverOption, 
    loadWin, unloadWin; 
    onSelect = function() { 
    var idx = this.selectedIndex; 
    if (idx !== -1) { 
     if (unsetChild) {recoverOption();} 
     removeOption(this.options[idx].value); 
    } 
    }; 
    removeOption = function (value) { 
    unsetChild = selectTo.querySelector('[value="'+value+'"]'); 
    nextSibling = unsetChild.nextSibling; 
    selectTo.removeChild(unsetChild); 
    }; 
    recoverOption = function() { 
    nextSibling 
     ? selectTo.insertBefore(unsetChild,nextSibling) 
     : selectTo.appendChild(unsetChild); 
    nextSibling = unsetChild = null; 
    }; 
    loadWin = function() { 
    selectFrom = document.getElementById('locationFrom'); 
    selectTo = document.getElementById('locationTo'); 
    selectFrom.addEventListener('change',onSelect,false); 
    }; 
    unloadWin = function() { 
    selectFrom.removeEventListener('change',onSelect,false); 
    selectFrom = selectTo = nextSibling = unsetChild = null; 
    }; 
    window.addEventListener('load',loadWin,false); 
    window.addEventListener('unload',unloadWin,false); 
});

來源

2013-11-24 18:46:00 hanzo2001

我已經制定了一個jQuery的解決方案（在谷歌瀏覽器測試）：

<html> 
<head> 
<meta http-equiv="content-type" content="text/html; charset=UTF-8" /> 
<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/2.0.3/jquery.min.js"></script> 
<script type="text/javascript"> 
function jqyHideSelectedOption(objSelect) { 
    var strIDSeleted = objSelect.id; 
    var strID2Handle; 
    if(strIDSeleted == "locationFrom") 
    { 
    strID2Handle = "#locationTo"; 
    } 
    else 
    { 
    strID2Handle = "#locationFrom"; 
    } 
    strIDSeleted = "#" + strIDSeleted; 
    // 
    // show all options: ie, restore default settings. 
    // 
    jQuery(strIDSeleted + " option").show(); 
    jQuery(strID2Handle + " option").show(); 
    // 
    // hide some options in the corresponding select: 
    // 
    var selectedCity = jQuery(strIDSeleted).val(); 
    if(jQuery(strID2Handle).val() == selectedCity) { 
    jQuery(strID2Handle).val(null); 
    } 
    jQuery(strID2Handle + " option[value='" + selectedCity + "']").each(function() { 
    jQuery(this).hide(); 
    }); 
} 
</script> 
</head> 
<body> 
<?php 
// 
//... 
// 
$MethodLocationFrom = mysql_query(" 
    SELECT DISTINCT Location_From AS 'locationFrom' 
    FROM trip 
"); 
echo "<form action='confirmation.php' method='post'>"; 
echo "<select class = 'LocationFrom' size='1' name='locationFrom' id='locationFrom' onchange='jqyHideSelectedOption(this)'>"; 
while($check = mysql_fetch_array($MethodLocationFrom)) 
{ 
    echo "<option value='" . $check['locationFrom'] . "'>" . $check['locationFrom'] . "</option>"; 
} 
echo"</select>"; 

$MethodLocationTo = mysql_query(" 
    SELECT DISTINCT Location_To AS 'locationTo' 
    FROM trip 
"); 
echo "<select class = 'LocationTo' size='1' size='1' name='locationTo' id='locationTo' onchange='jqyHideSelectedOption(this)'>"; 
while($check = mysql_fetch_array($MethodLocationTo)) 
{ 
    echo "<option value='" . $check['locationTo'] . "'>" . $check['locationTo'] . "</option>"; 
} 
echo"</select>"; 
// 
//... 
// 
?> 
</body> 
</html>

對於從城市#locationFrom選擇，使用一個onchange = '（這）jqyHideSelectedOption'。在此選擇中選擇城市時，我們將此城市名稱隱藏在＃位置選擇列表中。如果我們選擇另一個城市，則隱藏城市再次顯示在目的地列表中，因此可再次用於目的地選擇。

如果選擇了目的地，該城市將隱藏在城市列表中。

我們不會從2選擇中刪除任何選項，而是暫時隱藏它，因爲如果選擇了另一個城市，該選項很有用，並且必須存在以供客戶選擇。

來源

2013-11-24 19:29:02 jacouh

這有效，但它只適用於最初的選項，當我更改locationTo選項框（第一個框）locationFrom選項框（第二個框）不更新，如何保持它每次更新？ –

這已經是一步了。 – jacouh

我更新了代碼以雙向工作。 – jacouh

回答

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