有没有办法将查询传递给php函数
我希望这个问题没有被问到其他地方,因为我看了,但我看了。我编写这个函数是为了为自己创建表。它可以工作,但我遇到的问题是,每次我调用这个函数时,它都会运行查询。是否有方法调用此函数并将查询传递给该函数有没有办法将查询传递给php函数,php,mysql,Php,Mysql,我希望这个问题没有被问到其他地方,因为我看了,但我看了。我编写这个函数是为了为自己创建表。它可以工作,但我遇到的问题是,每次我调用这个函数时,它都会运行查询。是否有方法调用此函数并将查询传递给该函数 function createawardtables($awardname, $awardshortname, $maxawards, $id){ $query="SELECT * FROM awards WHERE id = $id"; $result = mysql_qu
function createawardtables($awardname, $awardshortname, $maxawards, $id){
$query="SELECT * FROM awards WHERE id = $id";
$result = mysql_query($query) or die("There was a problem with the SQL query: " . mysql_error());
while($row = mysql_fetch_array($result)){
$order = array("","1st","2nd","3rd","4th","5th","6th","7th","8th",'9th',"10th");
echo "<table><th colspan=4><font color=maroon size='4pt'><u><b>Orders of the $awardfullname</b></u></font></th>";
for($i=1; $i<$maxawards+1; $i++) {
${$awardshortname.$i} = dateconvert(($row["$awardshortname$i"]), 2);
${$awardshortname.$i.'by'} = $row["$awardshortname{$i}by"];
${$awardshortname.$i.'memo'} = $row["$awardshortname{$i}memo"];
echo "<tr>
<td>$order[$i] Order given on </td>
<td><input type=text name='$awardshortname$i' title='mm/dd/yyyy' value= '${$awardshortname.$i}' size=8/></td>
<td>by <input type=text name='$awardshortname{$i}by' title='Name of who gave the award.' value='${$awardshortname.$i.'by'}'size=15/></td>
<td>for <input type=text name='$awardshortname{$i}memo' title='What the award was give for.' value='${$awardshortname.$i.'memo'}'size=15/></td>
</tr>";
if($i==10 or $awardshortname=="initiate"){
echo "</table><br />";
};
};
};
mysql_free_result($result);
};
函数createawardtables($awardname、$awardshortname、$maxawards、$id){
$query=“从id=$id的奖励中选择*”;
$result=mysql\u query($query)或die(“SQL查询有问题:”.mysql\u error());
while($row=mysql\u fetch\u数组($result)){
$order=数组(“,”1“,”2“,”3“,”4“,”5“,”6“,”7“,”8“,”9“,”10”);
回显“$awardfullname订单”;
对于($i=1;$i当然,只需在函数中添加一个新参数(如$awardname
),并在调用“mysql\u query()”时使用它
另外,强烈建议不要使用mysql.*
函数:
从PHP5.5.0开始,此扩展已被弃用,并将在中删除
相反,应该使用MySQLi或PDO_MySQL扩展
给你
function createawardtables($result, $awardname, $awardshortname, $maxawards, $id)
{
while($row = mysql_fetch_array($result)){
$order = array("","1st","2nd","3rd","4th","5th","6th","7th","8th",'9th',"10th");
echo "<table><th colspan=4><font color=maroon size='4pt'><u><b>Orders of the $awardfullname</b></u></font></th>";
for($i=1; $i<$maxawards+1; $i++) {
${$awardshortname.$i} = dateconvert(($row["$awardshortname$i"]), 2);
${$awardshortname.$i.'by'} = $row["$awardshortname{$i}by"];
${$awardshortname.$i.'memo'} = $row["$awardshortname{$i}memo"];
echo "<tr>
<td>$order[$i] Order given on </td>
<td><input type=text name='$awardshortname$i' title='mm/dd/yyyy' value= '${$awardshortname.$i}' size=8/></td>
<td>by <input type=text name='$awardshortname{$i}by' title='Name of who gave the award.' value='${$awardshortname.$i.'by'}'size=15/></td>
<td>for <input type=text name='$awardshortname{$i}memo' title='What the award was give for.' value='${$awardshortname.$i.'memo'}'size=15/></td>
</tr>";
if($i==10 or $awardshortname=="initiate"){
echo "</table><br />";
};
};
};
mysql_free_result($result);
};
您正在使用并且应该使用。您还(取决于$id
来自何处)容易受到现代API的攻击,这将使您更容易使用。感谢您的帮助我尝试了类似的方法,但我使用了call_user_func进行了尝试
$query="SELECT * FROM awards WHERE id = $id";
$result = mysql_query($query) or die("There was a problem with the SQL query: " . mysql_error());
createawardtables($result, $awardname, $awardshortname, $maxawards, $id);