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[英]selected value get from db into dropdown select box option using php mysql error
[英]PHP select option using value from MySQL
使用此代碼,我得到下拉菜單來選擇:
<select name='select'>
$sql = mysql_query("SELECT sport FROM kategorija");
mysql_close();
while ($result = mysql_fetch_array($sql)) {
<OPTION selected='{$kategorija}' VALUE='" . $result[0] . "'>" . $result[0] . "</OPTION>
}
</select></td>
這是我的桌子的樣子:
ID DropDownMenu xxx yyy zzz
如何設置下拉菜單選擇的值是與該ID連接的值。 就我而言,我總是從下拉菜單中選擇最后一個值。
試試這個:UPDATED
$out = '<select name="select">';
$sql = mysql_query("SELECT sport FROM kategorija");
mysql_close();
while ($result = mysql_fetch_assoc($sql)) {
$out .= "<OPTION selected='" .$result['$kategorija']. "' VALUE='" .$result[0]. "'";
if ($result['$kategorija'] == 'something') {
$out .= "selected";
}
$out .= ">" .$result[0]. "</OPTION>";
}
$out .= "</select></td>";
echo $out;
首先將所選運動在數據庫中存儲到變量中
例如:
$sport='cricket';//substitute with database fetch value
Sport: <select name="sport" >
<option value="All" >All</option>
<?php $a="select * FROM kategorija";
$d2=mysql_query($a);
while($d4=mysql_fetch_array($d2))
{?> <option value="<?php echo $d4['sport']; ?>"<?php if($sport==$d4['sport']){ echo "selected";} ?> > <?php echo $d4['sport']; ?> </option> <?php
}
?></select>
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